Solve Euler-Cauchy Problem: x^2y" - 2xy' +2y = x^4e^x

[ryan8888]

Homework Statement

Find the General Solution of x^2y" - 2xy' +2y = x^4e^x

Homework Equations

The Attempt at a Solution

I just recently began learning about the Euler-Cauchy Equation and have been able to follow it until I hit the part for the Complementary Equation. I'm having a little truoble understanding how to set up the Wronskian.

Here is what I came up with for this problem:

My auxilliary equation is M² + (a-1)m + b =0 with a = -2 and b= 2

So I have: m² - 3m +2 =0
Solving for the Roots: (m-1)(m+2) = 0
m₁ = 1 and m₂ =2

Because these are real and distinct roots my characteristic equation is y_c = c₁x + c₂x²

This is where I get stuck. I have to solve for the Particular solution and the example I found uses a Wronskian to do so. I'm familiar with the method in which I would make a guess at the general solution, take the derivatives and then substitute them into the original ode and solve for the constants.

Here I know the Wronskian is w(x,x²) But I'm not entirely sure where the x and x² are obtained from. Are they simply the 2 x components of the ODE and secondly if they are does the order in which they are listed have any bearing on the outcome of the solution?

 

[HallsofIvy]

You are talking about the "variation of coefficients" method which can be written as a single formula involving the Wronskian. Here are the more fundamental ideas behind that formula.

We know that solutions to the associated homogeneous equation are x and x^2 so we look for a solution of the form y= xu(x)+ x^2v(x) for some functions u and v. y'= u+ xu'+ 2xv+ x^2v'. Now, there are an infinite number of solutions of that form so we can simplify our search by adding the requirement that xu'+ x^2v'= 0 so that y'= u +2xv. Differentiating again, y''= u'+ 2v+ 2xv'.

Now, put those into the differential equation:
x^2y''- 2xy'+ 2y= x^2u'+ 2x^2v+ 4x^3v'- 2x^2u'- 2x^3v'+ 2xu+ 2x^2v
= 2x^3v'= x^4 e^x
v&#039;= \frac{1}{2}xe^{x}[/itex]<br /> (It is that &quot;2x^3&quot; that was the Wronskian.)<br /> You can integrate that to find v. And then the previous xu&amp;#039;+ x^2v&amp;#039;= xu&amp;#039;+ \frac{1}{2}x^3e^x= 0 reduces to u&amp;#039;= -\frac{1}{2}x^2e^x.<br /> <br /> <br /> You don&#039;t normally get a single equation in one variable as happened here. What typically happens is that you get an equation in both u&#039; and v&#039; (There is no u&#039;&#039; or v&#039;&#039; because of or requirement that xu&amp;#039;+ x^2v&amp;#039;= 0. There is no u or v because x and x^2 satisfied the homogeneous equation.) which, together with our &quot;requirement&quot;, gives two equations to solve for u&#039; and v&#039;.<br /> <br /> You may know that if you solve a pair of equations, like ax+ by= p, cx+ dy= q, the solutions will involve the determinant<br /> \left|\begin{array}{cc}a &amp;amp; b \\ c &amp;amp; d\end{array}\right|<br /> in the denominator.<br /> <br /> That is the Wronskian here.

 

[ryan8888]

Okay I see what you hac done there.

I was browsing around the internet because I'm not familiar with variation of parameters and found the following:

w = \left|\begin{array}{cc}y1a &amp; y2 \\ y&#039;1 &amp; y&#039;2\end{array}\right| which I solved to get x^2

The next step was to solve for w1 and w2:

w1 = \left|\begin{array}{cc}0 &amp; y2 \\ f(x) &amp; x^2\end{array}\right| which is:
w1 = \left|\begin{array}{cc}0 &amp; x^2 \\ x^2e^x &amp; 2x \end{array}\right| according to my example. Why is the f(x) = x^2e^x when the f(x) in the problem is x^4e^x?

I feel like I am missing a step in here.

Thanks for your continued help

You are talking about the "variation of coefficients" method which can be written as a single formula involving the Wronskian. Here are the more fundamental ideas behind that formula.

We know that solutions to the associated homogeneous equation are x and x^2 so we look for a solution of the form y= xu(x)+ x^2v(x) for some functions u and v. y&#039;= u+ xu&#039;+ 2xv+ x^2v&#039;. Now, there are an infinite number of solutions of that form so we can simplify our search by adding the requirement that xu&#039;+ x^2v&#039;= 0 so that y&#039;= u +2xv. Differentiating again, y&#039;&#039;= u&#039;+ 2v+ 2xv&#039;.

Now, put those into the differential equation:
x^2y&#039;&#039;- 2xy&#039;+ 2y= x^2u&#039;+ 2x^2v+ 4x^3v&#039;- 2x^2u&#039;- 2x^3v&#039;+ 2xu+ 2x^2v
= 2x^3v&#039;= x^4 e^x
v&#039;= \frac{1}{2}xe^{x}[/itex]<br /> (It is that &quot;2x^3&quot; that was the Wronskian.)<br /> You can integrate that to find v. And then the previous xu&amp;#039;+ x^2v&amp;#039;= xu&amp;#039;+ \frac{1}{2}x^3e^x= 0 reduces to u&amp;#039;= -\frac{1}{2}x^2e^x.<br /> <br /> <br /> You don&#039;t normally get a single equation in one variable as happened here. What typically happens is that you get an equation in both u&#039; and v&#039; (There is no u&#039;&#039; or v&#039;&#039; because of or requirement that xu&amp;#039;+ x^2v&amp;#039;= 0. There is no u or v because x and x^2 satisfied the homogeneous equation.) which, together with our &quot;requirement&quot;, gives two equations to solve for u&#039; and v&#039;.<br /> <br /> You may know that if you solve a pair of equations, like ax+ by= p, cx+ dy= q, the solutions will involve the determinant<br /> \left|\begin{array}{cc}a &amp;amp; b \\ c &amp;amp; d\end{array}\right|<br /> in the denominator.<br /> <br /> That is the Wronskian here.