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solve the equation $(x-1)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=(x-1)^2$ given that $x$ and $e^x$ are the particular integrals of the equation without the right hand member. What is the right hand member here? $(x-1)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=(x-1)^2$ |
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It means that $x$ and $e^x$ are solutions of the homogeneous associated equation: $$(x-1)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0 $$ The right hand side is $(x-1)^2$ |
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Divide both sides of this equation by $x-1$: $$y''-\frac{x}{x-1} y' + \frac{1}{x-1} y = x-1$$ The right-hand side is $x-1$. Multiply through by $e^{-x}/(x-1)$. This is an integrating factor which puts the equation in a Sturm-Liouville form: $$\frac{d}{dx} \left [\frac{e^{-x}}{x-1} y'\right] + \frac{e^{-x}}{(x-1)^2} y = e^{-x}$$ |
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Just solve it by variation of parameters. $y_1=x$ , $y_2=e^x$ The Wronskian of these two functions is $\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=(x-1)e^x$ $\therefore y=C_1x+C_2e^x-x\int\dfrac{1}{(x-1)e^x}e^x(x-1)~dx+e^x\int\dfrac{1}{(x-1)e^x}x(x-1)~dx=C_1x+C_2e^x-x\int~dx+e^x\int xe^{-x}~dx=C_1x+C_2e^x-x^2-e^x\int x~d(e^{-x})=C_1x+C_2e^x-x^2-x+e^x\int e^{-x}~dx=C_1x+C_2e^x-x^2-1$ |
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