$\displaystyle y''+4y'+3y=e^{-t}$, given $\displaystyle y(0)=y'(0)=1$

My Attempt:

Taking Laplace transforms on both sides $\displaystyle $ $\displaystyle [s^2\bar y-sy(0)-y'(0)]+4[s\bar y-y(0)]+3\bar y=\frac{1}{s+1} $

$\displaystyle [s^2+4s+3]\bar y=\frac{1}{s+1}+s+5 $

$\displaystyle \bar y=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$

Resolving into partial fractions,

$\displaystyle \frac{A}{s+1}+\frac{B}{(s+1)^2}+\frac{C}{s+3}=\frac{s^2+6s+6}{(s+1)(s^2+4s+3)}$

I get A=7/2; B=1/2 and C=1

Taking inverse, $\displaystyle y=\frac{7}{2}e^{-t}+\frac{1}{2}te^{-t}+e^{-3t}$

The given answer is $\displaystyle y=\frac{7}{4}e^{-t}-\frac{1}{2}te^{-t}-\frac{3}{4}e^{-3t}$

I can't find where I am going wrong. Please help.

  • $\begingroup$ Everything is correct up to the evaluation of A, B and C. Their values are correct. $\endgroup$ Jul 23, 2014 at 15:59

1 Answer 1


The value of $C$ is wrong.

It should be $A=\frac{7}{4}, B=\frac{1}{2}, C=-\frac{3}{4}$


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