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Method of Reduction of Order and Variation of Parameters

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

One of my friends gave me the following question. I am posting the question and the answer here so that he could check his work.

Question:

This question concerns the differential equation,

\[x\frac{d^{2}y}{dx^2}-(x+1)\frac{dy}{dx}+y=x^2\]

and the associated homogeneous differential equation,Wronskian - Wikipedia, the free encyclopedia

\[x\frac{d^{2}y}{dx^2}-(x+1)\frac{dy}{dx}+y=0\]

a) Show that \(y_{1}(x)=e^x\) is a solution of the homogeneous differential equation.

b) Use the method of reduction of order to show that a second linearly independent solution of the homogeneous differential equation is, \(y_{2}(x)=x+1\).

Hint:
\(\int xe^{-x}\,dx=-(x+1)e^{-x}+C\)


c) Use the method of variation of parameters to find the general solution of the given non-homogeneous differential equation.

Hint:
Write the differential equation in standard form and remember the hint from part (b).
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
For part a),

Substitute \(y=e^x\) in the right hand side of,

\[x\frac{d^{2}y}{dx^2}-(x+1)\frac{dy}{dx}+y=0\]

and show that it satisfies the above equation.

For part b),

Let \(y_{2}(x)=v(x)e^x\) be the second linearly independent solution. Substituting this in the homogeneous equation we get,

\[x\frac{d^{2}}{dx^2}[v(x)e^x]-(x+1)\frac{d}{dx}[v(x)e^x]+v(x)e^x=0\]

Simplification yields,

\[\Rightarrow xe^xv''(x)+(2xe^x-(x+1)e^x)v'(x)+(xe^x-(x+1)e^x+e^x)v(x)=0\]

\[\Rightarrow xe^xv''(x)+(xe^x-e^x)v'(x)=0\]

\[\Rightarrow xv''(x)+(x-1)v'(x)=0\]

Using separation of variables,

\[v(x)=A(x+1)e^{-x}+B\]

where \(A\) and \(B\) are arbitrary constants. Therefore,

\[y_{2}(x)=A(x+1)+Be^x\]

Since the second term, \(Be^x\) and \(y_{1}(x)\) are linearly dependent we can neglect that term to get, \(y_{2}(x)=x+1\) as the second solution to the homogeneous differential equation. The two solutions \(y_{1}(x)=e^x\) and \(y_{2}(x)=x+1\) are linearly interdependent which could be verified from the Wronskian.

\[W(y_1,\,y_2)(x)=\begin{vmatrix} e^x & x+1 \\ e^x & 1 \end{vmatrix}=-xe^x\neq 0\]

Therefore the general solution of the homogeneous part is,

\[y_{c}(x)=A(x+1)+Be^x\]

For part c),

I am sure you must have learnt about the variation of parameters method and I am not going to go through the details. A step by step analysis of the variation of parameters method can be found >>here<<. Our differential equation is,

\[x\frac{d^{2}y}{dx^2}-(x+1)\frac{dy}{dx}+y=x^2\]

\[\Rightarrow \frac{d^{2}y}{dx^2}-\left(\frac{x+1}{x}\right)\frac{dy}{dx}+\left( \frac{1}{x}\right)y=x\]

Let, \(q(x)=-\frac{x+1}{x},\, r(x)=\frac{1}{x}\) and \(g(x)=x\). Now the particular solution to the non-homogeneous differential equation is,

\[y_{p}(x)=-y_1(x)\int\frac{y_2(x)g(x)}{W(y_1,\,y_2)(x)}\,dx+y_{2}(x)\int\frac{y_1(x)g(x)}{W(y_1,\,y_2)(x)}\,dx\]

Substitute and simplify to get,

\[y_{p}(x)=-(x^2+2x+2)\]

Therefore the general solution of,

\[x\frac{d^{2}y}{dx^2}-(x+1)\frac{dy}{dx}+y=x^2\]

is,

\[y(x)=y_{p}(x)+y_{c}(x)=-(x^2+2x+2)+A(x+1)+Be^x\]

where \(A\) and \(B\) are arbitrary constants.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
I have a comment that can really put parts (b) and (c) together. Use reduction of order on the whol DE. Using

$y = u e^x$ gives

$xu'' +(x-1)u' = x^2 e^{-x}\;\;\;(1)$

Letting $v = u'$ make (1) a first order linear DE - easy to integrate. Then $v = u'$ gives a separable equation to integrate. I still don't know why this technique is not as well publicized as should be.