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I'm trying to solve the following differential equation:

$y'' + y = x \cdot e^x$

I already have the homogeneous solution: $y_h = c_1 \cdot cos(x) + c_2 \cdot sin(x)$ but I'm struggeling to find the particular solution. I tried using $y_p = a \cdot x \cdot e^x$ but that didn't quite work, or maybe I did something wrong...

Can somebody explain to me how to solve this?

Thanks.

Edit: Here's the solution:

$y_p = (a \cdot x + b) \cdot e^x$

$y_p' = (a + a \cdot x + b) \cdot e^x$ and $ y_p'' = (2 \cdot a + a \cdot x + b) \cdot e^x$

$y_p'' + y_p = (2 \cdot a \cdot x + 2 \cdot a + 2 \cdot b) \cdot e^x$

$\Rightarrow 2 \cdot a = 1 \Rightarrow a = \frac{1}{2}$ and $2 \cdot a + 2 \cdot b = 0 \Leftrightarrow 2 \cdot a = 1 = - 2 \cdot b \Rightarrow b = - \frac{1}{2}$

$\Rightarrow y_p = (a \cdot x + b) \cdot e^x = \frac{x-1}{2} \cdot e^x$

All in all: $y = y_h + y_p = c_1 \cdot cos(x) + c_2 \cdot sin(x) + \frac{x-1}{2} \cdot e^x$

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Try $y_p = (ax+b)e^x$.

[Additional characters to bypass the 30 characters rule]

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    $\begingroup$ oh.. yeah, that works, thanks $\endgroup$ – zutru Jan 21 at 17:11

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