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MAT2442018S => MAT244Tests => Quiz1 => Topic started by: Mariah Stewart on January 25, 2018, 04:51:29 PM

Find the general solution of the given differential equation, and use it to determine how solutions as t approaches infinity.
ty' + 2y = sin(t), t>0
Answer: Find an integrating factor

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Find the general solution of the given differential equation, and use it to determine how solutions behave as t→∞
ty′+2y=sin(t), t>0

$$ty'+2y=sint$$
First, we divide both sides of the given equation by $t$, we get: $\\$
$$y'+{2\over t}y={sint\over t}$$
Now the differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={2\over t}$ and $g(t)={sint\over t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{2\over t}dt}}=e^{2lnt}=e^{lnt}\cdot e^{lnt}=t\cdot t=t^{2}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^2y'+2ty=tsint$$
and $$(t^2y)'=tsint$$
Integrating both sides:
$$\int{(t^2y)'}=\int{tsint}$$
Thus, $$t^2y=\int{tsint}$$
For $\int{tsin(t)}$, we use Integration By Parts:$\\$
Let $u=t, dv=sint$.$\\$
Then $du=dt, v=cost$$\\$
Hence, $$\int{tsint}=uv\int{vdu}$$
$$\int{tsint}=tcost\int{costdt}$$
$$\int{tsint}=tcost+\int{costdt}$$
$$\int{tsint}=tcost+sint+c$$
Thus $$t^2y=tcost+sint+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $t^2$, we get the general solution:
$$y={(sinttcost+c)/t^2}$$
Since given $t>0$, $y\rightarrow 0$ as $t \rightarrow \infty$.

$$ty'+2y=sint$$
First, we divide both sides of the given equation by $t$, we get: $\\$
$$y'+{2\over t}y={sint\over t}$$
Now the differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={2\over t}$ and $g(t)={sint\over t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{2\over t}dt}}=e^{2lnt}=e^{lnt}\cdot e^{lnt}=t\cdot t=t^{2}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^2y'+2ty=tsint$$
and $$(t^2y)'=tsint$$
Integrating both sides:
$$\int{(t^2y)'}=\int{tsint}$$
Thus, $$t^2y=\int{tsint}$$
For $\int{tsin(t)}$, we use Integration By Parts:$\\$
Let $u=t, dv=sint$.$\\$
Then $du=dt, v=cost$$\\$
Hence, $$\int{tsint}=uv\int{vdu}$$
$$\int{tsint}=tcost\int{costdt}$$
$$\int{tsint}=tcost+\int{costdt}$$
$$\int{tsint}=tcost+sint+c$$
Thus $$t^2y=tcost+sint+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $t^2$, we get the general solution:
$$y={(sinttcost+c)/t^2}$$
Since given $t>0$, $y\rightarrow 0$ as $t \rightarrow \infty$.

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