# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-1 => Topic started by: Mariah Stewart on January 25, 2018, 04:51:29 PM

Title: Q1-TUT0501
Post 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

Title: Re: Q1-TUT0501
Post by: Mariah Stewart on January 25, 2018, 04:52:52 PM
Here's a clearer image
Title: TUT 0501
Post by: Zihan Wan on January 25, 2018, 04:54:59 PM
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
Title: Re: Q1-TUT0501
Post by: Meng Wu on January 25, 2018, 06:22:17 PM
$$ty'+2y=sint$$
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}$$\\ First, we find the integrating factor \mu(t) \\ As we know. \mu(t)=\exp^{\int{p(t)dt}} \\ Thus, \mu(t)=\exp^{\int{{2\over t}dt}}=e^{2ln|t|}=e^{ln|t|}\cdot e^{ln|t|}=t\cdot t=t^{2}$$\\$
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={(sint-tcost+c)/t^2}$$Since given t>0, y\rightarrow 0 as t \rightarrow \infty. Title: Re: TUT 0501 Post by: Meng Wu on January 25, 2018, 06:22:29 PM$$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^{2ln|t|}=e^{ln|t|}\cdot e^{ln|t|}=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:
$$y={(sint-tcost+c)/t^2}$$
Since given $t>0$, $y\rightarrow 0$ as $t \rightarrow \infty$.
Title: Re: TUT 0501
Post by: Victor Ivrii on February 02, 2018, 07:47:18 AM