# Find the solution of the given initial value problem. y'+(2/t)y=(cos t)t^2 , y(3.14)=0, t>0

Find the solution of the given initial value problem.
${y}^{\prime }+\left(2/t\right)y=\left(\mathrm{cos}t\right){t}^{2},y\left(3.14\right)=0,t>0$
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repotasonwf
Given equation is
${y}^{1}+\frac{2}{t}y=\frac{\mathrm{cos}t}{{t}^{2}}\to \left(1\right)$
This is a linear equation of first order in y
where $P\left(t\right)=\frac{2}{t}andQ\left(t\right)=\frac{\mathrm{cos}t}{{t}^{2}}$
$\int P\left(t\right)dt=\int \frac{2}{t}dt=2\mathrm{log}t=\mathrm{log}{t}^{2}$
Integrating Factor $={e}^{\int P\left(t\right)dt}={e}^{\mathrm{log}{t}^{2}}={t}^{2}$
Multiplying (1) with ${t}^{2}$, we get
${t}^{2}{y}^{1}+2ty=\mathrm{cos}t\to \left(2\right)$
(2) can be written as
$\frac{d}{dt}\left[y{t}^{2}\right]=\mathrm{cos}t$
$\int \frac{d}{dt}\left[y{t}^{2}\right]dt=\int \mathrm{cos}tdt$
$⇒y\ast {t}^{2}=\mathrm{sin}t+C$
$⇒y\left(t\right)=\frac{\mathrm{sin}t}{{t}^{2}}+\frac{C}{{t}^{2}}$
we know that $y\left(\pi \right)=0$
so $\frac{\mathrm{sin}\pi }{{\pi }^{2}}+\frac{C}{{\pi }^{2}}=0⇒C=0$
Therefore the general solution of (1) is $y=\frac{\mathrm{sin}t}{{t}^{2}}$