Solving differential equation using method of undetermined coefficients?

Question

Solve the differential equation $$y'' + 2y' = 100$$

I know the solution should be the homogeneous equation + the particular solution. I got the homogeneous equation to be $y_h= C_1+C_2e^{-2t}$ but I can't figure out the particular solution. When I tried it, the $y''$ and $2y'$ both were 0 making it $0 = 100$ which is wrong, so I'm not sure where to go. I saw that the particular solution online was $50t$ but I'm not sure how they got that. The initial conditions are $y(0)=1$ and $y'(0)=2$ if they help.

Answer 1

Solve the differential equation $y'' + 2y' = 100$ For the particular solution simply try $y'_p=A$ or if you prefer $y_p=At+B$. You get $A=50$ as expected. So that $y_p=50t+B$. And the constant B is absorbed by the constant you already have $C_1$. Finally you get $y(t)=C_1+C_2e^{-2t}+50t$

Answer 2

The ODE is lacking an $y$ term. Then it may be simpler to solve with $z:=y'$,

$$z'+2z=100.$$

The homogeneous solution is $z=ce^{-2t}$, and a particular solution is given by a constant, because the derivative of a constant vanishes. Hence you easily conclude

$$y'=z=ce^{-2t}+50$$ and by integration,

$$y=c_1e^{-2t}+50t+c_2.$$

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You can determine the constant in the end, or as you go,

$$y'(0)=z(0)=c+50=2$$ so that $c_1=\frac{-48}{-2}$, and

$$y(0)=24+c_2=1.$$

Finally,

$$y(t)=24e^{-2t}+50t-23.$$

Answer 3

Let $100=2c$ and let $y'=w$ $$y''+2y'=2c \implies w'+2(w-c)=0$$ let $w-c=z$, then $$z'+2z=0 \implies z=C_1 e^{-2x}\implies w=C_1 e^{-2x}+c\implies y'=C_1 e^{-2x}+c$$ $y'(0)=2$ gives $C_1=-48$. $$\implies y=C_1\frac{e^{-2x}}{-2}+50x+C_2$$ Then $y(0)=1$ gives $C_2)=-23$ Finally we have $$y(x)=24e^{-2x}+50x-23$$