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Evgesh-ka [11]
3 years ago
5

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Mathematics
1 answer:
Ad libitum [116K]3 years ago
3 0
First you'll need the complementary solutions. The characteristic equation for this ODE is

D^2+2D+1=(D+1)^2=0\implies D=-1

with multiplicity 2. This means two linearly independent solutions will be y_1=e^{-x} and y_2=xe^{-x}.

Via variation of parameters, the particular solution will take the form

y_p=y_1u_1+y_2u_2

where

u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx

The Wronskian is

W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}

So, you have

u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx
u_1=-\displaystyle\int x\log x\,\mathrm dx
u_1=\dfrac14x^2-\dfrac12x^2\log x

and

u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx
u_2=\displaystyle\int\log x\,\mathrm dx
u_2=x(\log x-1)

So the solution to the ODE is

y=C_1y_1+C_2y_2+y_1u_1+y_2u_2
y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}
y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)
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<h3><u>Explamation </u><u>:</u></h3>

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