Find the general solution of the given differential equation

$\displaystyle y''+y=7sin(2t)+5tcos(2t)$

Since g(t) is a sum I'll have 2 equations.

$\displaystyle y''+y=7sin(2t)$

$\displaystyle y''+y=5tcos(2t)$

I'm kind of confused on which forms of Y to use.

Would $\displaystyle Y_1 (t)=Asint(2t)$ and $\displaystyle Y_2 (t)=(At+B)sin(2t)+(At+B)cos(2t)$ ?

Re: Find the general solution of the given differential equation

No, you don't need to do two equations.

Have you evaluated the homogeneous equation's solution yet?

Re: Find the general solution of the given differential equation

Do you mean $\displaystyle y(t)=y_c (t)+ Y(t)$ ?

If so, $\displaystyle y(t) = c_1 cos(t)+c_2 sin(t) + Y(t)$

I can't find $\displaystyle Y(t)$ though.

Re: Find the general solution of the given differential equation

For "linear differential equations with constant coefficients" if the "right hand side" functions are those that we expect as solutions to a homogeneous equation, $\displaystyle e^{ax}$, $\displaystyle sin(x)$, $\displaystyle cos(x)$, powers of t, and products of those, then you can expect such functions as solutions of the equation.

Here, the right side is 7sin(2t)+ 5t cos(2t). So try a solution of the form y= Asin(2t)+ Bcos(2t)+ Ctsin(2t)+ Dtcos(2t).

Then y'= 2A cos(2t)- 2B sin(2t)+ Ct cos(2t)+ C sin(2t)- Dt sin(2t)+ D cos(2t)= (Ct+ 2A+ D)cos(2t)+ (C-2B- Dt) sin(2t) and

y''= C cos(2t)- 2(Ct+ 2A+ D)sin(2t)- D sin(2t)+ 2(C- 2B- Dt) cos(2t).

Put those into the differential equations to get equations to solve for A, B, C, and D.

Re: Find the general solution of the given differential equation

I was able to get that A=11/9, B=0, C=0, and D=-5/3.

But based off of my answer choices A should equal either 1/9 or -7/9.

Re: Find the general solution of the given differential equation

HallsofIvy made a calculation error or two.

$\displaystyle \begin{align*}y(t) & = A\sin(2t) + B\cos(2t) + Ct\sin(2t) + Dt\cos(2t) \\ y'(t) & = 2A\cos(2t) - 2B\sin(2t) + 2Ct\cos(2t) +C\sin(2t) -2Dt\sin(2t) + D\cos(2t) \\ & = (2A + D)\cos(2t) + (C-2B)\sin(2t) + 2Ct\cos(2t) - 2Dt\sin(2t) \\ y''(t) & = -2(2A+D)\sin(2t) + 2(C-2B)\cos(2t) -4Ct\sin(2t)+2C\cos(2t) -4Dt\cos(2t) -2D\sin(2t) \\ & = (4C-4B-4Dt)\cos(2t)-(4A+4D+4Ct)\sin(2t) \\ y(t) + y''(t) & = (4C-3B-3Dt)\cos(2t) - (3A + 4D + 3Ct)\sin(2t) \end{align*}$

So $\displaystyle 5t = 4C-3B-3Dt$ and $\displaystyle 7 = -(3A+4D+3Ct)$. Hence $\displaystyle B = C = 0$, $\displaystyle D = -\dfrac{5}{3}$, and $\displaystyle A = -\dfrac{1}{9}$.

That's closer to what you were expecting...

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Re: Find the general solution of the given differential equation

Thanks to you all.

So do I ever need to have 2 or more individual equations when g(t) is a sum of equations to find the particular solution? I have a picture to show what I mean.

Re: Find the general solution of the given differential equation

One of these days I really have to learn arithmetic!

Re: Find the general solution of the given differential equation

You can solve them all at once or one at a time, however you prefer.

Re: Find the general solution of the given differential equation

Ok, thanks again to you all