Melissa's question at Yahoo! Answers regarding solving a linear first order ODE

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MarkFL

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We are given to solve the first order linear ODE:

$\displaystyle \frac{df}{dy}+f(y)=\sin(2y)$

I am assuming $\displaystyle f$ is the dependent variable and $\displaystyle y$ is the independent variable, and that the (t) is a typo.

We may begin by calculating our integrating factor $\displaystyle \mu(y)$:

$\displaystyle \mu(y)=e^{\int\,dy}=e^y$

Multiply the ODE by the integrating factor:

$\displaystyle e^y\frac{df}{dy}+f(y)e^y=e^y\sin(2y)$

Now, we may rewrite the left side as the differentiation of a product:

$\displaystyle \frac{d}{dy}\left(e^yf \right)=e^y\sin(2y)$

Integrate with respect to $\displaystyle y$:

$\displaystyle \int\frac{d}{dy}\left(e^yf \right)\,dy=\int e^y\sin(2y)\,dy$

On the right side, we may use integration by parts:

$\displaystyle u=\sin(2y)\,\therefore\,du=2\cos(2y)\,dy$

$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$

and so we may state:

$\displaystyle I=\int e^y\sin(2y)\,dy=e^y\sin(2y)-2\int e^y\cos(2y)\,dy$

Now, using integration by parts again:

$\displaystyle u=\cos(2y)\,\therefore\,du=-2\sin(2y)\,dy$

$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$

and we have:

$\displaystyle I=e^y\sin(2y)-2\left(e^y\cos(2y)+2\int e^y\sin(2y)\,dy \right)$

$\displaystyle I=e^y\sin(2y)-2e^y\cos(2y)-4I$

Solve for $\displaystyle I$:

$\displaystyle I=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$

Now, back to integrating the ODE, we have:

$\displaystyle e^yf=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$

Hence, the general solution is given by:

$\displaystyle f(y)=\frac{\sin(2y)-2\cos(2y)}{5}+c_1e^{-y}$

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