Question 1187735
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'\ +\ y\ =\ 5\sin(2x)]


First-order linear ODE:  *[tex \Large y'(x)\ +\ p(x)y\ =\ q(x)]


Let *[tex \Large \mu(x)\ =\ e^{\small{\int\Large \,p(x)\,dx}}\ =\ e^{\small{\int\Large \,1\,dx}}\ =\ e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^x\frac{dy}{dx}\ +\ e^xy\ =\ 5e^x\sin(2x)]


Note that *[tex \Large e^x\ =\ \frac{d}{dx}e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^x\frac{dy}{dx}\ +\ \frac{d}{dx}e^xy\ =\ 5e^x\sin(2x)]


Use the reverse product rule: *[tex \Large f\frac{dg}{dx}\ +\ g\frac{df}{dx}\ =\ \frac{d}{dx}(fg)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}(e^xy)\ =\ 5e^x\sin(2x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{d}{dx}(e^xy)\,dx\ =\ \int\,\5e^x\sin(2x)\,dx]


Left side:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{d}{dx}(e^xy)\,dx\ =\ e^xy]


Right side:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\5e^x\sin(2x)\,dx]


Integrate by parts twice: *[tex \Large \int\,f\,dg\ =\ fg\ -\ \int\,g\,df]


Let *[tex \Large f\ =\ 5\sin(2x)] and *[tex \Large g\ =\ e^x]


Then *[tex \Large df\ =\ 10\cos(2x)dx] and *[tex \Large dg\ =\ e^xdx], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\5e^x\sin(2x)\,dx\ =\ 5e^x\sin(2x)\ -\ \int\,e^x\,10\cos(2x)dx\ +\ C]


Integrate by parts a second time to obtain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\5e^x\sin(2x)\,dx\ =\ 5e^x\sin(2x)\ -\ 10e^x\cos(2x)\ -\ 4\int\,5e^x\sin(2x)\,dx\ +\ C]


Then add *[tex \Large 4\int\,5e^x\sin(2x)\,dx] to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\int\,\5e^x\sin(2x)\,dx\ =\ 5e^x\sin(2x)\ -\ 10e^x\cos(2x)\ +\ C]


And divide by 5


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\5e^x\sin(2x)\,dx\ =\ e^x\sin(2x)\ -\ 2e^x\cos(2x)\ +\ C]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^xy\ =\ e^x\sin(2x)\ -\ 2e^x\cos(2x)\ +\ C]


Divide by *[tex \Large e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sin(2x)\ -\ 2\cos(2x)\ +\ C]


You did not give initial conditions with this one, so there is only a general solution.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

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