# How to solve the DE

#### idontknow

$$\displaystyle y''-y =-e^{-x}$$.

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#### idontknow

My approach using Wronskian : $$\displaystyle dW(y,e^x )=-dx \;$$ ; $$\displaystyle \; W(y,e^x )=C_1 -x \;$$ ; $$\displaystyle \; y'e^x -e^{x} y=C_1 -x$$.

$$\displaystyle y'-y=C_1 e^{-x} - xe^{-x} \;$$ ; $$\displaystyle \; (ye^{-x} )'=C_1 e^{-2x} -xe^{-2x} \;$$ ; $$\displaystyle \; ye^{-x} = C_2 -\dfrac{C_1 }{2} e^{-2x}+ \dfrac{\left(2x+1\right)\mathrm{e}^{-2x}}{4}$$.
$$\displaystyle y=C_2 e^{x} -\dfrac{C_1 }{2}e^{-x} + \dfrac{\left(2x+1\right)\mathrm{e}^{-x}}{4}$$.

#### v8archie

Math Team
It's a second order linear ODE for which the homogeneous equation has the characteristic polynomial $r^2-1=0$ so the complementary solution is $y_c = c_1 e^{-x} + c_2 e^{x}$. The particular solution, by the method of undetermined coefficients, then has the form $y_p=(Ax+B)e^{-x}$.

idontknow

#### skipjack

Forum Staff
I'll use $\text{A}$ and $\text{B}$ for the constants of integration.

Multiply by $e^{-x}$, then integrating gives $e^{-x}y' + e^{-x}y = \frac12e^{-2x} + 2\text{A}$.

Multiply by $e^{2x}$, then integrating gives $e^xy = \frac12x + \text{A}e^{2x} + \text{B}$.

Hence $y = \frac12xe^{-x} + \text{A}e^x + \text{B}e^{-x}$, which is consistent with idontknow's solution.

idontknow