Question

Solve by Laplace transform:
\[\frac{dy}{dx}+3y=10sint\] y(0)=0

Answer 1

:D

Answer 2

\[sY - 0+3Y = \frac{10}{s^2+1}\]\[Y = \frac{10}{(s^2+1)(s+3)}\]\[\frac{As+B}{s^2+1}+\frac{C}{s+3}=\frac{10}{(s^2+1)(s+3)}\]
A +C = 0 -(1)
3A +B =0 -(2)
3B+C =10 -(3)
(3) - (1) = 3B-A = 10 -(4)
Solve (2) and (4): A = -1 B =3
C = -A = 1
\[Y = \frac{-s+3}{s^2+1}+\frac{1}{s+3}=\frac{-s}{s^2+1}+\frac{3}{s^2+1}+\frac{1}{s+3}\]\[y=-cost + 3sint + e^{-3t}\] ......

Answer 3

yes, thats correct.

Answer 4

Ha! That smiley face helps :D