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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

Solve the differential equation

$x'=x+\sin(t)$

ok this uses x rather than y which threw me off

so rewrite as

$x'-x=\sin(t)$

thus $u(t)=e^{-t}$

$e^{-t}x'-e^{-t}x=e^{-t}\sin{t}$

and

$(e^{-t}x)'=e^{-t}\sin{t}$

intergrate thru

$\displaystyle e^{-t}x=\int {e^{-t} \sin(t)} dt = -1/2 e^{-t}\sin(t) - 1/2 e^{-t} \cos(t) +e^{-t} c$

divide thru

$\displaystyle x(t)=-\frac{\sin t}{2}-\frac{\sin t}{2}+ce^t$

ok well typos probably

W|A returned $\displaystyle x(t) = c_1 e^t - \frac{\sin(t)}{2} - \frac{cos(t)}{2}$

$x'=x+\sin(t)$

ok this uses x rather than y which threw me off

so rewrite as

$x'-x=\sin(t)$

thus $u(t)=e^{-t}$

$e^{-t}x'-e^{-t}x=e^{-t}\sin{t}$

and

$(e^{-t}x)'=e^{-t}\sin{t}$

intergrate thru

$\displaystyle e^{-t}x=\int {e^{-t} \sin(t)} dt = -1/2 e^{-t}\sin(t) - 1/2 e^{-t} \cos(t) +e^{-t} c$

divide thru

$\displaystyle x(t)=-\frac{\sin t}{2}-\frac{\sin t}{2}+ce^t$

ok well typos probably

W|A returned $\displaystyle x(t) = c_1 e^t - \frac{\sin(t)}{2} - \frac{cos(t)}{2}$

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