Verify that the function $y(x)=C_1e^{-2x}+C_2e^x$ is a solution of the linear equation $\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=0$ for any choice of constants $C_1$ and $C_2$. In addtion determine the constants for the initial conditions $y(0)=2$, $y'(0)=1$.
So first I took the derivatives:
$y'(x)=-2C_1e^{-2x}+C_2e^x$
$y''(x)=4C_1e^{-2s}+C_2e^x$
When I plugged the above into $\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=0$ I found that it was equal to $y(x)=C_1e^{-2x}+C_2e^x$.
Trying to find the constants I got:
For $y(0)$, $C_1=2$ and $C_2=0$
For $y'(0)$, $C_1=\frac{-1}{2}$ and $C_2=0$
Is this correct?
$\endgroup$ 31 Answer
$\begingroup$I suppose that they wanted you to check that $$\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=0\tag1$$ if $$y=C_1e^{-2x}+C_2e^x$$ As, you did, compute the expressions of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ and plug all expressions (including $y$) in $(1)$; the result is $0$ so the given $y$ is a solution.
Now, from the equations, apply the conditions $$2=C_1+C_2$$ $$1=-2C_1+C_2$$ Solve for $C_1,C_2$ to get the particular solution.
All of what you did is very correct; I think that you could present the different steps in a slightly different manner to better show the steps. A few more words could help to show how you did proceed.
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