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##### Quiz-4 / TUT0801 Quiz4
« on: October 18, 2019, 04:44:43 PM »
Consider the initial value problem,

$y^{\prime \prime}+2 y^{\prime}+5 y=4 e^{-t} \cos 2 t, y(0)=1,$ $y^{\prime}(0)=0$

This is a non-homogeneous differential equation. Then its general solution is given by, $y=$complimentary solution $+$ particular solution.

For the complimentary solution, consider the homogeneous equation

$y^{\prime \prime}+2 y^{\prime}+5 y=0$

Then its characteristic equation is given by,

$r^{2}+2 r+5=0$

Then,

$r=\frac{-2 \pm \sqrt{4-20}}{2}$

$r=\frac{-2 \pm 4 i}{2}$

$r=-1 \pm 2 i$

Therefore, the complimentary solution is,

$y_{c}(t)=c_{1} e^{-t} \cos 2 t+c_{2} e^{-t} \sin 2 t$

where $c_1$ and $c_2$ are arbitrary constant.

For the particular solution, by the method of undetermined co-efficients, suppose $Y(t)=A t e^{-t} \cos 2 t+B t e^{-t} \sin 2 t$ is a function which satisfies the equation $Y^{\prime \prime}+2 Y^{\prime}+5 Y=4 e^{-t} \cos 2 t$

Since $Y=A t e^{-t} \cos 2 t+B t e^{-t} \sin 2 t$

Then $Y^{\prime}=A e^{-t} \cos 2 t-A t e^{-t} \cos 2 t-2 A t e^{-t} \sin 2 t+B e^{-t} \sin 2 t -B t e^{-t} \sin 2 t+2 B t e^{-t} \cos 2 t$

$=(A-A t+2 B t) e^{-t} \cos 2 t+(-2 A t+B-B t) e^{-t} \sin 2 t$

And

$Y^{\prime \prime}=(2 A-3 A t+4 B-4 B t) e^{-t} \cos 2 t+(-4 A+4 A t-2 B-3 B t) e^{-t} \sin 2 t$

Substitute $Y,Y'$ and $Y''$ in $Y^{\prime \prime}+2 Y^{\prime}+5 Y=4 e^{-t} \cos 2 t$ to get,

$(-2 A-3 A t+4 B-4 B t+2 A-2 A t+4 B t+5 A t) e^{-t} \cos 2 t +(-4 A+4 A t-2 B-3 B t-4 A t+2 B-2 B t+5 B t) e^{-t} \sin 2 t=4 e^{-t} \cos 2 t$

$4 B e^{-t} \cos 2 t-4 A \sin 2 t=4 e^{-t} \cos 2 t$

$4 B \cos 2 t-4 A \sin 2 t=4 \cos 2 t$

On equating coefficients on both sides,

$B=1$ and $A=0$

Then, the particular solution is,

$Y(t)=t e^{-t} \sin 2 t$

So, the general solution of $y^{\prime \prime}+2 y^{\prime}+5 y=4 e^{-t} \cos 2 t, y(0)=1,$ $y^{\prime}(0)=0$ is,

$y=y_{c}(t)+Y(t)$

$y=c_{1} e^{-t} \cos 2 t+c_{2} e^{-t} \sin 2 t+t e^{-t} \sin t$

Differentiate this equation with respectt to $t$, to get,

$y^{\prime}=-c_{1} e^{-t} \cos 2 t-2 c_{1} e^{-t} \sin 2 t-c_{2} e^{-t} \sin 2 t+2 c_{2} e^{-t} \cos 2 t +e^{-t} \sin 2 t-t e^{-t} \sin 2 t+2 t e^{-t} \cos 2 t$

Use the initial condition $y(0)=1$, implies,

$1=c_1+0+0$

$c_1 = 1$

Use the initial condition $y'(0)=0$, implies,

$0=-c_{1}+2 c_{2}$

$c_{2}=\frac{1}{2}$

Hence the required solution of given initial value problem is,

$y=e^{-t} \cos 2 t+\frac{1}{2} e^{-t} \sin 2 t+t e^{-t} \sin 2 t$

2
##### Quiz-3 / TUT0801 Quiz3
« on: October 11, 2019, 02:05:09 PM »
Find the Wronskian of the given pair of functions.

$$\cos (t), \sin (t);$$

Suppose $y_{1}(t)=\cos t$, $y_{2}(t)=\sin t$

Then Wronskian for this pair is given by

$W\left(y_{1}, y_{2}\right)=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|$

$=\left|\begin{array}{cc}{\cos t} & {\sin t} \\ {-\sin t} & {\cos t}\end{array}\right|$

$=\cos ^{2} t+\sin ^{2} t$

$=1$

i.e. $W=1$

3
##### Quiz-2 / TUT0801 Quiz2
« on: October 04, 2019, 02:04:47 PM »
$$x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3}$$

First, let's show the given DE $x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0$ is not exact.

Define $M(x, y)=x^{2} y^{3}, N(x, y)=x\left(1+y^{2}\right)$

$$M_{y}=\frac{\partial}{\partial y}\left[x^{2} y^{3}\right]=3 x^{2} y^{2}$$

$$N_{x}=\frac{\partial}{\partial x}\left[x\left(1+y^{2}\right)\right]=1+y^{2}$$

Since $3 x^{2} y^{2} \neq 1+y^{2}$, this implies the given DE is not exact.

Now, let's show that the given DE multipled by the integrating factor $\mu(x, y)=\frac{1}{x y^{3}}$ is exact.

That is to show

$$\frac{1}{x y^{3}} x^{2} y^{3}+\frac{1}{x y^{3}} x\left(1+y^{2}\right) y^{\prime}=x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$$

is exact.

Define $M^{\prime}(x, y)=x, N^{\prime}(x, y)=y^{-3}+y^{-1}$

Since

$$M_{y}^{\prime}=\frac{\partial}{\partial y}(x)=0$$

$$N_{x}^{\prime}=\frac{\partial}{\partial x}\left[y^{-3}+y^{-1}\right]=0$$

By theorem in the book, we can conclude that $x+\left(y^{-3}+y^{-1}\right) y^{\prime}=0$ is exact.

Thus, we know there exists a function $\phi(x, y)=C$ which satisfies the given DE.

Also,

$$\frac{\partial \phi}{\partial x}=x$$

$$\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$$

Integrate $\frac{\partial \phi}{\partial x}=x$ with respect to $x$ we have

$$\phi(x, y)=\frac{1}{2} x^{2}+g(y)$$

Take derivative on both sides with respect to $y$ we get

$$\frac{\partial \phi}{\partial y}=g^{\prime}(y)$$

Since we know that $\frac{\partial \phi}{\partial y}=y^{-3}+y^{-1}$

Then $g^{\prime}(y)=y^{-3}+y^{-1}$

Integrate with respect to $y$ we have

$g(y)=-\frac{1}{2} y^{-2}+\ln |y|+C$

Altogether, we have $\phi(x, y)=\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$, which means

$$\frac{1}{2} x^{2}-\frac{1}{2} y^{-2}+\ln |y|=C$$

is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0 \forall x$ is also a solution to the given DE.

4
##### Quiz-1 / TUT0801 Quiz 1
« on: September 28, 2019, 11:18:38 AM »
$$t^{3} y^{\prime}+4 t^{2} y=e^{-t}, \quad y(-1)=0, \quad t<0$$
First divide by $t^3$ on both side of the equation, we get

$$y^{\prime}+\frac{4}{t} y=\frac{e^{-t}}{t^{3}}$$

Using the method of integrating factor we have equation for  $u(t)$

$$u(t)=e^{\int \frac{4}{t} d t}=e^{4 \ln (t)} =t^{4}$$

Multiply both sides by $t^{4}$

$$t^{4} y^{\prime}+4 t^{3} y=t e^{-t}$$

$$\left(t^{4} y\right)^{\prime}=t e^{-t}$$

$$t^{4} y=\int t e^{-t}$$

$$t^{4} y=-t e^{-t}-e^{-t}+C$$

$$y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}+\frac{C}{t^{4}}$$
to check $C,$ plug in condition $y(-1)=0$

$$y(-1)=e-e+C=C=0$$
Plug in $C=0$ gets

$$y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}$$

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