Category Archives: Study Guide

Improved Euler’s Method

Posted on May 17, 2015 by | 11 comments

 

As we proceed through the course, we are usually given a first-order differential equation that could be solved. However, there are a lot of problems that cannot be solved. The first order equations could be divided into the linear equation, separable equation, nonlinear equation, exact equation, homogeneous equation, Bernoulli equation, and non-homogeneous equations. However, most of the separable and exact equation cannot always be presented the solution in an explicit form. It’s hard to find the value for a particular point in the function. There are some of the equations that do not fall into any of the categories above. So we introduce the method called Euler’s Method. We will be able to use it to approximate the solutions to a differential equation. In the Euler method, we will be given a differential equation which is the slope of a function, and define a step size for the integral ( the smaller steps sizes you have, the more accurate approximation values you will be get ). We generate a new point by starting at an initial point, we plug in this point into the given function, this will be the slope of the initial point. Then, then next new point will be the y_0 plus  step size h time the previously calculated slope. the general formula is,y_(n+1)=y_n+hf(t_n,y_n) However, the error for the Euler’s Method depends on the step size. the only way to decrease the error is to reduce the step size, but it will increase the amount of calculations. it only roughly decreases the error by half.

 

Now, we introduce an improved Euler’s Method. This method is quite similar to the Euler’s method. In the Euler’s Method we approximate the function by a rectangular shape (see graph below):

 

rectangular shape (see graph below):However, this approximate does not include the area that under the curve. To improve the approximation, we use the improved Euler’s method.The improved method, we use the average of the values at the initially given point and the new point. We define the integral with a trapezoid instead of a rectangle. The trapezoid has more area covered than the rectangle area. It will also provide a more accurate approximation.

improve

 

It is hard to predict the solution curve is concave up or concave down in reality. The ideal prediction line would exactly hit the curve at next predict point. The Euler’s Method generates the slope based on the initial point, and we don’t know if the next point will be on this slope line, unless we use a computer to plot the equation. Sometimes, we might overestimate the value or underestimate the value. The Improved Euler’s Method addressed these problems by finding the average of the slope based on the initial point and the slope of the new point, which will give an average point to estimate the value. It also decreases the errors that Euler’s Method would have.

The improved Euler’s Method simply divided into three steps as following:

Steps in Improved Euler’s Method:

Step 1 find the  k_1=f(t_n,y_n)

Step 2 find the k_2=f(t_n+h,y_n+hk_1)

Step 3: find y_(n+1)=f(t_n+h,y_n+h((k_1+k_2)/2))

 Given a first order linear equation y’=t^2+2y, y(0)=1, estimate  y(2), step size is 0.5. 

new point 1

point 2

point 3 new

point 4

 

Summary summary 1

 

Note: it is very important to write the (t_n,y_n)and t_(n+1) at the beginning of each step because the calculations are all based on these values. This method involved with a lot of calculations, it is recommended after each point, write the values in a table. It will be easy for yourself to look up and check. For each point, the calculations approach to the next new point are the same, so if you set up the three steps, it will be very clear for you to continue to the next step.

I think this video is pretty helpful, and make a clear point on the improved Euler’s Method and a example include in the video. please check out this video.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Definition of Laplace Transform

Posted on May 13, 2015 by | 7 comments

Laplace transform is a method used to go from one domain to another domain. In this case we go from the time domain (t) to the frequency domain  (s).

As for most conversions, if there is one way to go from one unit to another, there should be a way to go backwards. It applies in this case also, although it isn’t necessarily a unit. In this case it’s simply called “Inverse Laplace Transform”. In this case we go from the Frequency domain (s), back to the time domain (t).

The notation for the Laplace transform is:

\mathcal{L}\{f(t)\}

The “L” is used to denote Laplace transform. What is inside the curly braces is the function you want to transform from the time domain to the frequency domain.

You may sometimes also see this notation which means the same as the one before:

fs

As for the Laplace transform, it is denoted:

Capture

This will give you F(s).

In order to get to the original time domain, you need to take the inverse Laplace transform of  F(s) which is:

eq0001MP

Now that we got the notation down for the Laplace transform we can go into more depth.

The general formula for the Laplace transform where ‘t’ is greater than or equal to zero:

F(s) =\int_0^{\infty} e^{-st} f(t)\, dt. We evaluate at t is greater than or equal to zero because we want to satisfy two conditions:

1. The function f(t) has to be piecewise continuous from the interval [0,A]. Simply means a function that is broken apart into different pieces but still continues on. For example:

piecewise-continuous

 

2. The same function f(t) must be of exponential order. This means that the function f(t) must be smaller than or equal to ke^(at) when t is grater than or equal to M. In this case the variables K, M, and a are just constants and K, M are positive.

As for the inverse Laplace transform, there isn’t any set equation or method of doing it. The way the inverse Laplace transform is denoted, is by the following:

laplace-transformation

It simply mean to get the function f(t) you would need to take the inverse Laplace transform of F(s).

The main reason we use Laplace transform is because it makes certain (not all) differential equations easier.

A small introduction on the steps to take when solving a Laplace transform problem. There are five steps that we can use to solve a differential equation using Laplace transform:

1. Have a differential equation to solve

2. Take the Laplace transform of both sides in the equation. This will give you a simple algebraic equation to solve.

3. Solve the algebraic equation

4. Simplify the algebraic so you have what you are solving for on the left side and what it is equal to on the right side. If you can simplify the right side it will make it easier. Once simplified use partial fractions to solve for the unknowns.

5. Take the inverse Laplace transform and you will have your solution for the differential equation.

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Repeated Roots Second Order Linear Homogeneous Equation

Posted on May 1, 2015 by | 4 comments

Overview

Second order linear equations become homogeneous when the linear function of y and y’ (which can be written in the form y” + p(t)y’ + q(t)y = g(t)) is equal to zero. In other words when g(t)=0.

This guide will be discussing how to solve homogeneous linear second order differential equation with constant coefficient, which is written in the following form:

y”+by’+cy = 0

The first step is to use the equation above to turn the differential equation into a characteristic    equation. The characteristic equation is written in the following form:

r2 +br+c = 0

Second to find the roots, or r1 and r2 you can either factor or use the quadratic formula:

 r2 = ± -b √b²-4ac

 2a

It is important to remember when to the particular equation above. There will problems where the variable a is not needed in the quadratic formula because there will be no a in the differential equation. In the cases where there is no a variable limit the a variable from the quadratic equation. The quadratic equation will the look like the following:

r2 = ± -b √b²-4c

2

Once you get repeated roots , or r1 and r2 from the characteristic equation then     y = ert is considered a solution of the differential equation.

The next step would be to plug r1 and r2 into the general equation:

y = C1 er1t +C2 er2t

Another important thing to realize and remember is that when solving a homogeneous equation for a repeated root the solution will end up cancelling out. In order to avoid this a “t” needs to be place in the general solution. The general solution for the repeated root will then be in the following form:

y = C1 er1t +C2t er2t

Sample Problem

Problem 1:

y”+12y’+36 = 0

Step 1: Turn the differential equation into a characteristic equation

r2 + br + c = 0

r2 + 12r + 36 = 0

Step 2: Factor the characteristic equation

r2 + 12r + 36 = 0

(r + 6) (r + 6) = 0

r1 = -6 r2 = -6

Step 3: Use y = ert as a solution for r1 and r2

y = ert

y = e-6t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = ert

y = e-4t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = C1 e-4t +C2t e-4t

y = C1 e-6t +C2t e-6t

 

Problem 2:

y”+8y’+16 = 0 , y(0) = 2   y’(0)= 6

Step 1: Turn the differential equation into a characteristic equation

r2 + br + c = 0

r2 + 8r + 16 = 0

Step 2: Factor the characteristic equation

r2 + 8r + 16 = 0

(r + 4) (r + 4) = 0

r1 = -4 r2 = -4

Step 3: Use y = ert as a solution for r1 and r2

y = ert

y = e-4t

Step 4: Plug r1 and r2 into the general solution for the repeated roots

y = C1 er1t +C2t er2t

y = C1 e-4t +C2t e-4t

The following three videos are form Khan Academy. I find these videos ever useful . I would recommend you watch them if you are still confused  about Repeated Roots of Second order linear homogeneous equation.

 

 

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Non-homogeneous Second Order Differential Equations Using Methods Of Undetermined Coefficients

Posted on May 1, 2015 by | 1 comment

Non-homogeneous differential equations are the same as homogeneous differential equations, However they can have terms involving only x, (and constants) on the right side. The interesting part of solving non homogeneous equations is having to guess your way through some parts of the solution process.
You also can write non-homogeneous differential equations in this format: y” + p(x)y’ + q(x)y = g(x).

The Reason I’ve chosen this problem is because it basically touches every aspect of a Non-homogeneous second order differential Equation using methods of undetermined coefficients.

y” + 6y’ + 9y = -578 sin 5t

The first step when dealing with undetermined or constant coefficients is getting the Characteristic equation. Which will later become the first half of our solution.

y” + 6y’ + 9y = 0
r^2 + 6r + 9 = 0
( r+3 ) ( r+3) = 0
R1 = -3 R2 = -3

When solving the characteristic equation sometimes we run into repeated roots. In this case we must use the following formula y = C1e^r1(t)+C2 (t)e^r2(t)

y (t) = C1e^-3t + C2te^-3t

After finding the characteristic equation our next step in this linear equation is to guess the y. In this case our y is going to include both sin and cosine. We make this guess because the cosine is going to come up eventually when finding the derivative and by adding it into the equation, it will allow us to cancel it out later on throughout the steps in the solution process. We also will need to find the y’ and y”.

y = C sin (5t) + D cos (5t)
y’ = 5 C cos (5t) – 5 D sin (5t)
y” = -25 C sin (5t) + 9 D cos (5t)

Now we plug it back into its original equation.

-25 C sin (5t) – 25 D cos (5t)
+30 C cos (5t) – 30 D sin (5t)
+9 C sin (5t) + 9 D sin (5t)
= -578 sin 5(t)

At this point we want to cancel and group any like terms.

(-16C -30D) sin (5t) + (-16D + 30C) cos (5t) = -578 sin (5t)

Resulting two equations.

(-16) (C) – (30) (D) = -578
(30) (C) – (16) (D) = 0
Now we must find Find C & D
(30) (C) – (16) (D) = 0
+16 (D) +16 (D)
C = (16/30) (D)

Now we plug in C back into the next equation.

(-16)(16/30D) – (30) (D) = -578
(-256/30)(D) – (30) (D) = -578
(-578/15)D = -578
D = 1514304785053481528808497

When calculating the (-256/30)(D) – (30) (D).
We use the calculator to find the fraction value by entering math > frac > enter.
Plug in D

C = (16/30) (D)
C = (16/30) (15)
C = 8

After finding all values we’re finally ready to plug in all variables and combine the homogeneous equation with its general solution.
Resulting in:
y (t) = C1e^-3(t) + C2(t)e^-3(t) + 8 sin 5(t) + 15 cos 5(t)
If we were given the case of finding a particular solution, we will have to take this a few steps further to and plug in the (t) value. But fortunately for us this general solution will suffice.

Below I have included videos that has helped me understand how to solve Non-homogeneous second order differential Equations using methods of undetermined coefficients. Khan academy has been an incredible help in the understanding of all things in relation to differential equations.

 

Problem listed above was taken from the midterm review, question 9 by Professor JReitz.

 

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Exact Equations

Posted on May 1, 2015 by | 1 comment

An exact equation is a first order differential equation. The steps to solving an exact equation is to check if MY  is equal to Nx. to check if they are equal you have to derive by M by y and derive N by x. the next step is to Integrate M with respect to X and obtain  and have to make sure to add the h(y) or else you will get an error later on. next take the derivative of  what the step done before but with respect to y.next you have to it equal to N and make sure that h(y) was also derived in the previous step. next solve for h'(y). after solving h'(y) we now integrate h'(y) to find h(y). After this we plug in h(y) to to the step where we integrated M with respect to x and make sure the equation is equal to C.

Next I will show an example for Exact equations:

finding the general solution for 2xy-9x^2+(2y+x^2+1) \frac{dy}{dx}=0.

the first step is to derive 2xy-9x^2 with respect y and 2y+x^2+1 with respect to x.

For 2xy-9x^2 we get 2x and for 2y+x^2+1 we get 2x as well so they are both equal to each other.

the next step is to integrate M with respect to y.

\int 2xy-9x^2 dy.

after integrating you should have gotten x^2y-3x^3+h(y).

the next step is deriving what you just integrated with respect to y.

when deriving you should get x^2+h'(y).

next make x^2+h'(y) equal to N which is 2y+x^2+1.

x^2+h'(y)=2y+x^2+1

the next step would be to find h'(y).

h'(y) would be 2y+1.

next we integrate h'(y).

\int h'(y)= \int 2y+1

after integrating we get

h(y)=y^2+y.

after finding h(y) we plug it into the equation we got when we integrated M with respect to y.

x^2y-3x^3+h(y)

the equation should now be x^2y-3x^3+y^2+y=c.

here are  videos on how to do exact equations if you need a better understanding:

https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/exact-equations/v/exact-equations-example-1

Resources:

http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

class notes

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Exact Equations

Posted on May 1, 2015 by | 1 comment

Exact Equation is a method of solving first order differential equations. First order differential equations come in different forms and there are many different methods for solving. One example of a first order differential equation will generally contain constants, an initial parameter and most importantly containing only 1 variable. Exact equation question consist of all that IVP contains however there are 2 unknown variables. A hint that the question tells you to approach the question using exact equation is when the equation is stated as df=(df/dx) dx +(df/dy) dy

For example:   df=(8x+9x^2y^3)+(9x^3y^2+cos(y)y')

8x+9x^2y^3 as My and 9x^3y^2+cos(y) as Nx

The steps to take are:

1.Test for Exactness of the equation

2.Take derivative of My with respect to y

8x+9x^2y^3 dy = 27x^2y^2

3.Take derivative of Nx in respect to x

9x^3y^2+cos(y)dx = 27x^2y^2

If the derivatives come out to be equal then the problem can be solved with the method of exact equation.

4.Taking integration of My with respect to x

\int 8x+9x^2y^3 dx = 4x^2+3x^3y^3+h(y)

5. Derive the answer from the previous part with respect to y where x will be consider a constant

4x^2+3x^3y^3+h(y) dy = 9x^3y^2+h'(y)

6.Set it equal to Nx equal to the answer from previous part and use algebra to take out parts of the equation that are equal to each other in order to find h'(y).

9x^3y^2+h'(y) = 9x^3y^2+cos(y)

As we can see 9x^3y^2 will get crossed out from both side of the equation when it is subtracted from one side to the other leaving h'(y) equal to cos(y).

7.Solve for h(y) so it can be placed back into the equation of 2i. In order to solve we must take the integration of h'(y) with respect to y.

\int h'(y) = \int cos(y) h(y) = sin(y)+c

8.Place h(y) back into the equation at 2i and we will get :

4x^2+3x^3y^3+sin(y) = c

I have included some useful materials on how to solve exact equation. These video has helped me with my homework as well as teaching me the material clearly in order for me to answer the exact equation question correctly on the test; therefore I highly recommend checking these videos out.

First video: Exact Differential Equations from patrickJMT of Youtube.

Second video: Exact differential equations:how to solve from Dr Chris Tisdell

 

If you have any more questions about exact equations after watching the videos, please go meet with your professor to get additional help. Thank you!

Resources:

http://www.cliffsnotes.com/math/differential-equations/first-order-equations/exact-equations

Example taken from Professor Reitz Spring 2015 Class.

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Repeated Roots, Second Order Linear

Posted on May 1, 2015 by | Leave a comment

Overview:

In a regular second order linear homogeneous equation the solution is easy to solve for after learning it in class. The equation should usually be in the following form when solving:(homogeneous=0). Where A, B, and C are constants and the highest degree should be the 2nd.

Ay”+By+Cy=0

The next step would be to get the characteristic equation of the differential equation from above or manage to put it into this form:

r²+Br+C

From here we need to be able to solve for the characteristic equation. This can be done in two ways. Either by factoring the equation or if it is not factor-able you may use the quadratic equation.

After getting the roots of the equation; we will call them r1 and r2, you can now proceed to solving the equation. The general solution of the characteristic equation is:

y=Ae^(r1*t)+Be^(r2*t)

This case is true for normal characteristic equations where the roots are real and different from each other , but in the case of this topic we would like to look at repeated roots; this would be when r1=r2.

If we try to solve the homogeneous equation using normal means with the repeated root the answer would come up to zero; that is not the correct answer because the values we are trying to solve for get cancelled out.

So in order to solve a equation with repeated roots we need to do one thing which is add a “t” in front of one of the original terms in the solution. Meaning the new equation we need to plug in values to would be:

 y=Ae^(r1*t)+tBe^(r2*t)

This would the final equation we use in order to solve an equation with repeated roots where A and B are the constants we want to solve for.

Example/Sample Problem:

Now lets try solving for a sample problem with repeated roots.

y”+18y’+81y=0 ; y(0)=-5, y'(0)=48

So the first step is to set up the characteristic equation:

r²+18r+81

Now we need to factor or solve for the values of r1 and r2.

(r+9)*(r+9)

r1=-9, r2=-9

Now as we can see the roots are exactly the same or repeated roots. So we have to plug it into the equation which would look like this:

 y=Ae^(-9t)+tBe^(-9t)

The next step would be to solve for the derivative of y.

y’=-9Ae^(-9t)-9tBe^(-9t)+Be^(-9t)

Now that we have both y and y’ we can now plug in the inital values of y(0)=-5 and y'(0)=48 to solve for the values of A and B.

For y(o)>>>  -5=Ae^(-9*0)+0*Be^(-9*0)

-5=A

For y'(0)>>> 48=-9Ae^(-9*0)-9*0Be^(-9*0)+Be^(-9*0)

48=-9A+B

48=-9(-5)+B

B=3

So now that we solved for both A and B we can now put it back into the original equation.

 y=-5e^(-9t)+3te^(-9t)

This is the final answer for the repeated roots problem.

3. In case I wasn’t clear enough with the explanation or steps needed to solve the equation, or even if you just didn’t understand because I skipped some steps while solving the equation you can refer to these two YouTube links for more help. The videos could help to provide an extra problem that you can work out while watching the video. The YouTuber provides excellent visuals and a great explanation of how the problems should be solved.

(Disclaimer: Not owned by me, all credits go to “Khan Academy”)

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Second order linear homogeneous equations with constant coefficients

Posted on April 30, 2015 by | 1 comment

You have probably seen first order linear equations. Now we”ll move to second order linear equations. What makes it linear? A differential equation is linear if it is a linear function of the variables y, y’, y” and so on.

The standard form of the second order linear equation is

y” + p(t)y’ + q(t)y = g(t)

where p(t), q(t), and g(t) are constant coefficients.
There can also be a constant coefficient in front of the y”.
When g(t) = 0, we can call the differential equation homogeneous, otherwise it is a non-homogeneous.
We can write the second order linear homogeneous equation can be written as:

ay”+by’+cy=0

From that, we can write that characteristic equation:

ar^2+br+c=0

From the characteristic equation, we can find the roots. The roots can come out to be real roots, repeated roots, or complex roots. But we will only focus on real roots.
How do we know if the roots are real roots? We can find out by find out the discriminant. To find out the discriminant, we use b^2-4ac. For the roots to be real, it must follow this condition:

b^2-4ac > 0

After finding out the roots, we can plug it into the general solution which is:

y(t)=C1e^(r1*t)+C2e^(r2*t)

Now we can figure out the derivative of the general solution which we’ll use later.

y'(t)=(r1)C1e^(r1*t)+(r2)C2e^(r2*t))

With problems like these, we will have initial condition, y(0) and y'(0), which will be used to find C1 and C2 to determine the particular solution.
You plug in the initial to determine C1 and C2.
Once you determined C1 and C2, plug them back into the general solution to get the particular solution satisfying the initial condition.
The initial conditions gives us a unique solution.
Example 1:      y” + 5y’ + 6y = 0         y(0) = 2     y'(0) = 3
We start out with the characteristic equation which is:

r^2 + 5r + 5 = 0

Now find the roots anyway you feel comfortable with.
One way is by using the quadratic formula.
Now we got

r1 = (-2)
r2 = (-3)

We can plug it into the general solution

y(t)=C1e^(-2t)+C2e^(-3t)

Now that we have to figure out the derivative of general solution which will be:

y'(t)=(-2)C1e^(-2t)-3C2e^(-3t)

Now we can plug in the initial values which are y(0) = 2     y'(0) = 3
For the initial condition y(0) = 2:

2=C1e^(-2*0)+C2e^(-3*0)
2=C1+C2

For the initial condition y'(0) = 3:

3=(-2)C1e^(-2*0)-3C2e^(-3*0)
3=-2C1-3C2

Now that we have 2 equations, we can use them to solve for C1 and C2:

2=C1+C2
3=-2C1-3C2

Multiply the first equation by 2 so the C1 can cancel out.

4=2C1+2C2
3=-2C1-3C2
7=-1C2
C2=(-7)

Now that we know C2, we can plug it back into any of the two equations to find C1.

2=(-7)+C1
C1=9

Lastly we plug the C1 and C2 into the general solution to determine the particular solution.

y(t)=9e^(-2t)-7e^(-3t)

Videos to help with Second order linear homogeneous equations with constant coefficients:
This video is a transition from first order linear differential equations to second order linear differential equations.


This video goes into more detail when the roots are repeated or complex. It only goes up the general solution


In this video, he does a whole example using initial conditions.

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Study Guide for Separable Equation

Posted on April 30, 2015 by | 4 comments
  1. Overview

What are separable equation and how to solve it ?

A differential equation is considered separable if the two variables can be moved to opposite sides of the equation. This facilitates solving a homogenous differential equation, which can be difficult to solve without separation.We are now going to start looking at nonlinear first order differential equations. The first type of nonlinear first order differential equations that we will look at is separable differential equations

A separable differential equation is any differential equation that we can write in the following form.

Note that in order for a differential equation to be separable all the y‘s in the differential equation must be multiplied by the derivative and all the x‘s in the differential equation must be on the other side of the equal sign. Also Consider the equation . This equation can be rearranged to . Any equation that can be manipulated this way is separable. The equation is solved by integrating both sides, resulting in an implicit solution. If an initial condition is provided, you can solve the implicit solution for an explicit solution, and determine the interval of validity, the range of x where the solution is valid. The interval of validity must be continuous and must contain the x-value given in the original condition.

2. Example  

Lets solve a simple question about separable equation.  differential equation: 2(dy)/(dx)=(y(x+1))/(x).                                                                       First of all we have to separate the variables that we have to move all the xs to one side and all the ys to the other. so first step we do is multiply dx on both side so we can cancel the y on the right side so it will give us                              (2dy)=(y(x+1)dx)/(x) and then we divide y on both sides too, so we can cancel the y on the right side. This will give us (2dy)/(y)=(1+1/x)dx  . Now all the y in one side and x is in the other side so we can solve it now by taking the integral on both sides. Therefor

(integral) (2dy)/(y)=integral(1+1/x)dx

So use the answer is 2 ln y= x+ln x+C

3. Video

Here is some videos about separable equation that I copy form KHAN academy if you still confuse about separable equation please watch and I think It will help you a lot.

This is video explain what is separable equation, watch the video for the detail.

In this video, he is explain some example.

Again some example that how to solve the separable equation correctly.

Best of luck on final.

 

 

 

 

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Study Guide to Bernoulli Equations

Posted on April 30, 2015 by | 1 comment

Overview

What are Bernoulli Equations?

Well to put it simply Bernoulli Equations are first order differential equations. What sets Bernoulli Equations apart from other first order differential equations is that they are nonlinear first order differential equation. When we was first introduced to first order differential equations we learned that the standard form  was :

y’ +p(t)y = g(t) ,  y(to) = yo

What separates Bernoulli Equations from other first order equations is that in standard form, it is not equal to some function that is linear but one that has an exact solution. What this means is that their is some power that is raised to the right side of the equation which we shall call n and n cannot be equal to 0 or 1. this is what makes Bernoulli Equations nonlinear. Now we can change the form of a standard first order linear equation into a nonlinear first order equation:

y’ +p(t)y = q(t)y^(n)  y(to)=yo

The above equation is now the standard form for a Bernoulli equation. What we can now understand from this equation is that  p(t) and q(t) are functions that are continuous. Since n is a real number that has to be n> 0 and n>1 for this  to be non linear.

Now the question is how do we solve Bernoulli Equations?

Sample Problem:

take an equation such as t^(2)y’ + 2ty – y^(3) = 0

and change it by adding y^(3) on both sides and we get:

t^(2)y’ + 2ty = y^(3)

Next we divide by t^(2) to on both sides of the equation so we can get it into standard form.

y’ + 2/yt = y^(3)/t^(2)       <————-STANDARD FORM

Now we can go into the steps to solve this equation;

1. Divide by y^(n)

y’/y^(3) + 2/yt/y^(3) =  (y^(3)/t^(2))/y^(3)

we get: (y’ + 2/yt)/y^(3) = 1/t^(2)

2. We substitute v = y^(1-n)

Since y^(3)

we get:v = y^(1-3) = y^(-2)

now we can rearrange the formula to so we can substitute v into it by moving the y^(2) to the numerator and we get:

y’/y^(3) + 2y^(-2)/t = 1/t^(2)

y’/y^(3) + 2v/t = 1/t^(2) <——————-Substitute v for y^(2)

Before we move to the solve Step we need to take the Derivative of the substitution v = y^(-2)

d/dy y^(-2)

dv/dt = -2y^(-3) dy/dt

(-1/2)v’ = -2y’/y^(3)(-1/2)

Now we can substitute for y’/y^(3) with the above equations and we get:

(-1/2)v’ + (2/t)v = 1/t^(2) <—————– First order linear equation

next we use the method of intergrating factor which states

mu = e^ intergral of (-4/t) which is equal to 1/t^(4)

we then multiply both sides b\by Mu and get the equation:

(dv(t)/dt)/t^(4) + d/dt(1/t^(4))v = (-2/t^(6))

Next we intergrate both sides and we come to the final answer of

y(t) = +- Sqrt(5t)/sqrt(ct^(5) + 2)

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