UrbanXrisis
- 1,192
- 1
-3y''-2y'+y=-t^2+2t+2e^{-4t}
i am to find the particular solution to this.
i started with the non-exponential:
y=At^2+Bt
y'=2At+B
y''=2A
(-6A-2B)+(-4A+B)t+A(t^2)
-6A-2B=0,-4A+B=2,A(t^2)=-1
A=-1, B=3
i started with the exponential:
y=Ce^{-4y}
y'=-4Ce^{-4t}
y''=16Ce^{-4t}
-48Ce^{-4t}+8Ce^{-4t}+Ce^{-4t}=2Ce^{-4t}
C=-\frac{2}{39}
so I think the particular solution is:
y_p=-t^2+3t- \frac{2}{39}e^{-4t}
but this is wrong, not sure why, any ideas?
i am to find the particular solution to this.
i started with the non-exponential:
y=At^2+Bt
y'=2At+B
y''=2A
(-6A-2B)+(-4A+B)t+A(t^2)
-6A-2B=0,-4A+B=2,A(t^2)=-1
A=-1, B=3
i started with the exponential:
y=Ce^{-4y}
y'=-4Ce^{-4t}
y''=16Ce^{-4t}
-48Ce^{-4t}+8Ce^{-4t}+Ce^{-4t}=2Ce^{-4t}
C=-\frac{2}{39}
so I think the particular solution is:
y_p=-t^2+3t- \frac{2}{39}e^{-4t}
but this is wrong, not sure why, any ideas?