[tex]-3y''-2y'+y=-t^2+2t+2e^{-4t}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

i am to find the particular solution to this.

i started with the non-exponential:

[tex]y=At^2+Bt[/tex]

[tex]y'=2At+B[/tex]

[tex]y''=2A[/tex]

[tex](-6A-2B)+(-4A+B)t+A(t^2)[/tex]

[tex]-6A-2B=0,-4A+B=2,A(t^2)=-1[/tex]

[tex]A=-1, B=3[/tex]

i started with the exponential:

[tex]y=Ce^{-4y}[/tex]

[tex]y'=-4Ce^{-4t}[/tex]

[tex]y''=16Ce^{-4t}[/tex]

[tex]-48Ce^{-4t}+8Ce^{-4t}+Ce^{-4t}=2Ce^{-4t}[/tex]

[tex]C=-\frac{2}{39}[/tex]

so I think the particular solution is:

[tex]y_p=-t^2+3t- \frac{2}{39}e^{-4t}[/tex]

but this is wrong, not sure why, any ideas?

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# Homework Help: Particular solution for dif eq

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