MAT244-2018S > Quiz-1

Q1-TUT0501

**Mariah Stewart**:

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

**Mariah Stewart**:

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**Zihan Wan**:

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

**Meng Wu**:

$$ty'+2y=sint$$

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 $t \rightarrow \infty$.

**Meng Wu**:

$$ty'+2y=sint$$

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 $t \rightarrow \infty$.

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