# antiderivative of $\frac{dy}{dx}$

Find the antiderivative of $\frac{dy}{dx}$=$e^{2x}-x$ and $y=5$ when $x=0$

the antiderivative formula is
$e^{kx}dx$=$\frac{1}{k}e^{kx}+c$

I did it ,
$\frac{dy}{dx}$=$e^{2x}-x$ and $y=5$ when $x=0$

$\frac{dy}{dx}=\frac{1}{2}e^{2x}+c$

but the right answer is $y=\frac{1}{2}(e^{2x}-x^2+9)$

can you help me out? thanks

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You made two major errors. First, you found the antiderivative only of the exponential term and forgot all about the $-x$ term: you should have got $$y=\frac12e^{2x}-\frac{x^2}2+c\;.$$ Note that this is $y$, not $dy/dx$.

Secondly, you found all of the antiderivatives, but the problem requires you to find the specific antiderivative such that $y=5$ when $x=0$. Take the corrected general antiderivative,

$$y=\frac12e^{2x}-\frac{x^2}2+c\;,$$

and substitute $y=5$ and $x=0$ to get

$$5=\frac12e^0-\frac02+c=\frac12+c\;;$$

clearly we must have $$c=5-\frac12=\frac92\;,$$ and the unique antiderivative that satisfies the extra condition is

$$y=\frac12e^{2x}-\frac{x^2}2+\frac92=\frac12\left(e^{2x}-x^2+9\right)\;.$$

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+1 nice solution –  dato datuashvili Jun 26 '12 at 10:39