variation problem

**transgalactic** · December 8th 2010, 09:17 AM

$(1-x)y''+xy'-y=2(x-1)^2e^x$
how to solve it?

i was told to show first that x and e^x are a solution of some homogeneous equation
i got that it the homogeneous equation?

but the partivular solution is tough
y_p=(ax^2+bx+c)e^x

its hard to differentiate every time

**TheEmptySet** · December 9th 2010, 08:53 AM

Originally Posted by transgalactic

$(1-x)y''+xy'-y=2(x-1)^2e^x$
how to solve it?

i was told to show first that x and e^x are a solution of some homogeneous equation
i got that it the homogeneous equation?

but the partivular solution is tough
y_p=(ax^2+bx+c)e^x

its hard to differentiate every time

Unfortunately you cannot use the method of undermined coefficients when the ODE does not have constant coefficients. You will need to use the variation of parameters.
Variation of parameters - Wikipedia, the free encyclopedia

since you are given the solutions to the homogeneous problem, we can solve for the particular solution.

1st. Put equation is standard form $y''+\frac{x}{1-x}y'-\frac{1}{1-x}y=-2(x-1)e^{x}$

2st. Assume solution of the form $y_p=u_1(x)\cdot x+u_2(x)\cdot e^{x}$

3nd.Calculate the Wronskian

$W=\begin{vmatrix}x & e^x \\ 1 & e^{x} \end{vmatrix}=(x-1)e^{x}$

Now

$W_1=\begin{vmatrix}0 & e^x \\ -2(x-1)e^{x} & e^{x} \end{vmatrix}=2(x-1)e^{2x}$

and

$W_2=\begin{vmatrix}x & 0 \\ 1 & -2(x-1)e^{x} \end{vmatrix}=-2x(x-1)e^{x}$

Now $\displaystyle u_1'(x)=\frac{W_1}{W}=2e^{x} \implies u_1(x)= 2e^{x}$

and

Now $\displaystyle u_2'(x)=\frac{W_2}{W}=-2x \implies u_2(x)=-x^2$

Putting all of this together gives

$\displaystyle y_p=2xe^{x}-x^2e^{x}$

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