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I am new to differential equations. I tried to find a series solution for this equation, but I don't know how to solve it.

\begin{equation} y'=x e^y + \cos x \\y(0)=1 \end{equation}

Actually, the problem needs the coefficient of $x^3$ in Maclaurin series solution.

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I forgot to write $y(0)=1$. I added it now. –  Mohammad Javad Naderi Jan 11 '13 at 10:22
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up vote 3 down vote accepted

For a series solution, put $y = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$. Assuming the equation han an analytic solution, you can differentiate and otherwise manipulate this series at will.

Also, the standard Maclaurin expansion of $e^t$ gives

\begin{align} e^y &= e^{a_0}e^{y-a_0} \\ &= e^{a_0}\left( 1 + (a_1 x + a_2 x^2 + \cdots) + \frac12(a_1 x + a_2 x^2 + \cdots)^2 + \cdots\right) \end{align}

The reason why I rewrite $e^y$ as $e^{a_0}e^{y-a_0}$ is to get an argument of $\exp$ which is $0$ for $x=0$.

Keeping track of enough terms, we get $$ e^y = e^{a_0}\left( 1 + a_1 x + \big(a_2 + \frac{a_1^2}2\big)x^2 + \big( a_3 + a_1a_2 + \frac{a_1^3}6 \big)x^3 + \cdots\right) $$ Hence (using the Maclaurin expansion of $\cos x$), $$ xe^y + \cos x = 1 + Ax + \big( Aa_1 - \frac12\big)x^2 + A\big( a_2 + \frac12a_1^2\big)x^3 + \cdots$$ where $A=e^{a_0}$. On the other hand $$y' = a_1 + 2a_2x + 3a_3x^2 + \cdots$$ and the coefficients of these two series must match. We get a system of equations, the first three being $$ \begin{cases} a_1 = 1 \\ A = 2a_2 \\ A a_1 - \frac12 = 3a_3 \end{cases} $$

From these equations, $a_3 = \frac13A - \frac16$.

(The chance that I managed to write all of this with no errors is slim, however...)

Somewhat surprisingly, the differential equation can be solved explicitly. You can check that

$$y = \sin x - \ln \bigg( C - \int xe^{\sin x}\,dx \bigg)$$

solves the differential equation.

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I think it should be $a_3=\frac{1}{3}A-\frac{1}{6}$. –  Mohammad Javad Naderi Jan 11 '13 at 10:21
@MJNaderi Yes, you're right. I fixed the sign error. –  mrf Jan 11 '13 at 10:27
Thanks. And because $y(0)=1$, then $A=e$ and $a_3=\frac{1}{3}e - \frac{1}{6}$. –  Mohammad Javad Naderi Jan 11 '13 at 10:37
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First, expand to first order: $y=1+ax+o(x^2)$. The diff. eq. shows $$y'=a+o(x)=\left[o(x)\right]+\left[1+o(x)\right]\implies a=1$$ Now consider $e^y$ to the first order: $$e^y=e^{1+x+o(x^2)}=e\cdot e^x\cdot e^{o(x^2)}=e(1+x)+o(x^2)$$ Thus to second order, $$y'=x\left[e(1+x)+o(x^2)\right]+\left[1-\frac{x^2}{2}+o(x^4)\right]$$ Collecting terms shows $y'[x^2]=e-\frac{1}{2}$. Thus, $$y[x^3]=\frac{2e-1}{6}$$

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