Toronto Math Forum

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,

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

$$
x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3}
$$

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.