 # Find the solution of the given initial value problem. ty'+2y=sin t, y(3.14/2)=1, t>0 Emmanuel Pace 2022-07-26 Answered
Find the solution of the given initial value problem.
$t{y}^{\prime }+2y=\mathrm{sin}t,y\left(3.14/2\right)=1,t>0$
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Bubbinis4
Given equation is $t{y}^{1}+2y=\mathrm{sin}t$
Dividing by t, we get
${y}^{1}+\frac{2}{t}y=\frac{\mathrm{sin}t}{t}$
This is a linear equation of first order in y
where
$\int P\left(t\right)dt=2\int \frac{1}{t}dt=2\mathrm{log}t=\mathrm{log}{t}^{2}$
Integrating Factor $={e}^{\int P\left(t\right)dt}={e}^{\mathrm{log}{t}^{2}}={t}^{2}$
Multiplying (1) with ${t}^{2}$, we get
${t}^{2}{y}^{1}+2ty=t\mathrm{sin}t\to \left(2\right)$
(2) can be rewritten as
$\frac{d}{dt}\left[y{t}^{2}\right]=t\mathrm{sin}t$
$\int \frac{d}{dt}\left[y\ast {t}^{2}\right]dt=\int t\mathrm{sin}t.dt$
$⇒y\ast {t}^{2}=t\left(-\mathrm{cos}t\right)-\int 1\left(-\mathrm{cos}t\right)dt$
$⇒y{t}^{2}=-t\mathrm{cos}t+\mathrm{sin}t+C$
$⇒y\left(t\right)=\frac{\mathrm{sin}t}{{t}^{2}}-\frac{\mathrm{cos}t}{t}+\frac{C}{{t}^{2}}$
we know that $y\left(\frac{\pi }{2}\right)=1$
so $\frac{1}{{t}^{2}}+\frac{C}{{t}^{2}}=1⇒C+1={t}^{2}⇒C={t}^{2}-1$
Therefore the general solution of (1) is $y=\frac{\mathrm{sin}t}{{t}^{2}}-\frac{\mathrm{cos}t}{t}+\frac{{t}^{2}-1}{{t}^{2}}$

We have step-by-step solutions for your answer!