Nonhomogenous Eq. and the Particular Solution

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jjm5119

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Find the solution of y''+10y'+21y=147e^(0t) with y(0)=8 and y'(0)=9

so my r values are -7 and -3 giving me y=c1e^(-7t)+c2e^(-3t)

initial conditions give me c1+c2=8

then taking the derivative and using the 2nd initial condition is get -7c1-3c2=9

Solving for c1 and c2 I get c1=-33/4 and c2=65/4

now for the particular solution. Since the right side is 147e^0t i can treat the ride side as 147

therefore yparticular(t)=some constant to be determined, say A

yparticular'(t)=0 and yparticular''(t)=0. Subbing these into the original equation i get 21A=147, so A equals 7 and my yparticular=7

so my solution is (-33/4)e^(-7t)+(65/4)e^(-3t)+7. the problem is that this isn't right. Where am i going wrong?
 
You made an arithmetic mistake somewhere. Since you have the correct (general) solution to the homogeneous equation, it vanishes when you plug it in. When you substitute y=A, it comes through as 21A=147, or A=7.
 
I understand how to get A, as you can see I got it. Where is my mistake? I've been over finding those constant five times and I know they are right. Can you specifically tell me when I'm going wrong so I can make changes.
 
jjm5119 said:
Find the solution of y''+10y'+21y=147e^(0t) with y(0)=8 and y'(0)=9

so my r values are -7 and -3 giving me y=c1e^(-7t)+c2e^(-3t)

initial conditions give me c1+c2=8

then taking the derivative and using the 2nd initial condition is get -7c1-3c2=9

Solving for c1 and c2 I get c1=-33/4 and c2=65/4

now for the particular solution. Since the right side is 147e^0t i can treat the ride side as 147

therefore yparticular(t)=some constant to be determined, say A

yparticular'(t)=0 and yparticular''(t)=0. Subbing these into the original equation i get 21A=147, so A equals 7 and my yparticular=7

so my solution is (-33/4)e^(-7t)+(65/4)e^(-3t)+7. the problem is that this isn't right. Where am i going wrong?
Click to expand...

your particular solution is:

y = 147/21 = 7

So your total solution is:

\(\displaystyle y \, = \, C_1\cdot e^{-7t}\, + \, C_2 \cdot e^{-3t} + 7\)

Now solve for constants:

\(\displaystyle C_1\, + \, C_2 = 1\) .....and so on
 
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