- #1

karush

Gold Member

MHB

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2000

Find the general solution of the given differential equation

$\displaystyle

ty^\prime - 2y =\sin{t}, \quad t>0\\$

Divide thru by $t$

$\displaystyle y^\prime - \frac{2}{t}y =\frac{\sin{t}}{t}$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -

\frac{2}{t} \, dx =e^{2\ln{t}}=t^{-2}\\$

Multiply thru with $t^{-2}$

$t^{-2}y^\prime - 2 t y= t^{-2}\sin{t}\\$

Simplify:

$(t^{-2}y)'= t^{-2}\sin{t}\\$

Integrate:

$\displaystyle t^2y=\int t^{-2}\sin{t} dt =

2t\sin(t)-(t^2-2)\cos(t)+c_1\\$

Answer from textbook

$y=\color{red}

{(c-t\cos t + \sin t )/t^2} \quad

y \to 0 \textit{ as } t \to \infty$

ok somewhere I am not approaching the bk ans

Find the general solution of the given differential equation

$\displaystyle

ty^\prime - 2y =\sin{t}, \quad t>0\\$

Divide thru by $t$

$\displaystyle y^\prime - \frac{2}{t}y =\frac{\sin{t}}{t}$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -

\frac{2}{t} \, dx =e^{2\ln{t}}=t^{-2}\\$

Multiply thru with $t^{-2}$

$t^{-2}y^\prime - 2 t y= t^{-2}\sin{t}\\$

Simplify:

$(t^{-2}y)'= t^{-2}\sin{t}\\$

Integrate:

$\displaystyle t^2y=\int t^{-2}\sin{t} dt =

2t\sin(t)-(t^2-2)\cos(t)+c_1\\$

Answer from textbook

$y=\color{red}

{(c-t\cos t + \sin t )/t^2} \quad

y \to 0 \textit{ as } t \to \infty$

ok somewhere I am not approaching the bk ans

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