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I have a calculus exam in college tomorrow and I've been stuck on this question for over and hour and a half now. $$y''-2y'-3y=4\sin2x$$ I've searched for answers but the videos tend to generalise the formula and I can't understand the form of the answer.

Thank you for your time, Alex.

3 Answers 3

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With $z=y'-3y$, you can check that $y''-2y'-3y=z'+z$.

Then solve

$$z'+z=4\sin(2x).$$

This is done by multiplying by $e^x$, and

$$z'e^x+ze^x=(ze^x)'=4\sin(2x)e^x.$$

Then integrating,

$$ze^x=4\int\sin(2x)e^x dx=\frac45(\sin(2x)-2\cos x)e^x+C$$

and

$$z=4\int\sin(2x)e^x dx=\frac45(\sin(2x)-2\cos x)+Ce^{-x}.$$

Next, solve

$$y'-3y=\frac45(\sin(2x)-2\cos x)+Ce^{-x}.$$

Similary,

$$y'e^{-3x}-3ye^{-3x}=(ye^{-3x})'=\frac45(\sin(2x)-2\cos x)e^{-3x}+Ce^{-4x}$$

and

$$ye^{-3x}=\int\left(\frac45(\sin(2x)-2\cos x)e^{-3x}+Ce^{-4x}\right)dx=\\ \frac{4\cos(2x)-7\sin(2x)}{65}e^{-3x}+Ce^{-4x}+C'.$$

Finally

$$y=\frac{4\cos(2x)-7\sin(2x)}{65}+Ce^{-x}+C'e^{3x}.$$

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First, you have to find a basis of functions for the homogeneous equation: $$y''-2y'-3y=0$$

You take the polynomial $$p(t)=t^2-2t-3$$ which comes from substituting $y$ by $t^0$, $y'$ by $t^1$, $y''$ by $t^2$ (and so on) in your equation. Its roots are $t_1=-1$ and $t_2=3$, both real so we have that $e^{t_1x}$ and $e^{t_2x}$ are solutions of the equation. (If you obtain a single root $t_1$, you can take $e^{t_1x}$ and $xe^{t_1x}$).

So that would be $$y_1(x)=e^{-x} \qquad y_2(x)=e^{3x}$$ You only need a particular solution of the complete equation, that is, a solution $y_p$ of $$y''-2y'-3y=4\sin(2x)$$ and all the solutions of your equation would be $$c_1y_1(x)+c_2y(2)+y_p(x), \qquad c_1, c_2\in\mathbb R$$

Now, note that $4\sin(2x)$ is $Im(4e^{2xi})$, so you will need to find a particular solution of $$y''-2y'-3y=4e^{2xi}$$ and take the imaginary part.

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    I know after using the (-b formula) auxillary, I have to put it into the form of: $$yh = Ae^m + Be^m$$ – 2017-01-09
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    Actually could you just do a quick run-through on how you got $$e^-x$$ and $$e^3x$$ Never mind. I just implemented auxillary eqn. wrong :') – 2017-01-09
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    I'm assuming you don't need me to complete the answer? xD – 2017-01-09
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    [Image | Press Me](http://imgur.com/a/0VJWi) That's what I have on my sheet so far. :D – 2017-01-09
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    [What We Did In Class | Press Me](http://imgur.com/a/waNnz) That's what form we use in class – 2017-01-09
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    yes, actually is the same way more or less. The thing is, you need to write $sin(2x)$ as a complex exponentiation, because your equation is not going to have a solution of the form $ax^2+bx+c$. But it will have one of the form $(ax^2+bx+c)e^{2xi}$, and that way the exponentials cancel. – 2017-01-09
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    Yeah that problem is I don't know what to do with the Sin(2x) :/ I don't understand how to pull out the value so I could group the values by their polynomials. – 2017-01-09
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    I feel very f#*%ing dumb :D – 2017-01-09
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    All right, try solving the equation $$y″−2y′−3y=4e^{2xi}$$ as you did in class, but instead of putting $$y=Cx^2+Dx+E$$ write $$y=e^{2xi}(Cx^2+Dx+E)$$ I don't know if you are familiar with complex numbers. – 2017-01-09
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    I presume $$4e^(2xi)$$ is 4Sin(2x) simplified down into its exponent? – 2017-01-09
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    actually you have $$e^{xi} = \cos x + i \sin x$$ where $i=\sqrt{-1}$ If you don't know complex numbers I'm not sure you are required to do it this way – 2017-01-09
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    No we haven't went into complex numbers, I know what they are but we haven't used them. See I'm sorry for frying your head like this, I'm not too good with First Order Differential Equations and this is my pit fall :D – 2017-01-09
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    no problem :3 I can't think of a way to solve this without getting into complex numbers – 2017-01-09
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    [This Image](http://imgur.com/a/waNnz) shows exactly how we did it in class. The thing is, this question has an x^2 at the end not 4Sin(2x) :D – 2017-01-09
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    yes, that method works with polynomials, but it won't work with sines or cosines perhaps you can try letting $$y=a\cos(2x)+b\sin(2x)$$ and follow that method and see if it turns out well – 2017-01-09
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    I'll give that a shot. Thank you SO MUCH for investing time to help me :D – 2017-01-09
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    It's nothing~ actually, it should work, there is a solution like that – 2017-01-09
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Here's an alternate way which I learned Answer

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    Thank you for showing your solution. I just don't understand entirely how you used the matrices :D – 2017-01-09
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    I think the solution below is wrong, just by a factor of 4 in the yp solution. The method is called variation of parameters and I've always found it the most structured method. Do you know Cramer's Rule? – 2017-01-09
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    Yeah I'm away of Cramer's Rule :) I found somewhat of a good way that I understand. [Youtube Video](https://youtu.be/o9Qxq7Bd4aE?t=1366) – 2017-01-09
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    aware of Cramer's Rule* – 2017-01-09
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    Just be careful with the method of undetermined coefficients, it doesn't always work. If say you had h(x)=tanx you might get stuck. – 2017-01-09
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    Yeah I understand thank you. It's just our lecturers and IV's are super strict on the method we use in the exam but after let us use any method ^_^ – 2017-01-09