Differential Equations

The given differential equation is:

$\frac{dy}{dx}=\frac{y}{x}.....\left(I\right)$

Rearrange the tems of the equation $\left(I\right)$

$\frac{dy}{y}=\frac{dx}{x}.....\left(II\right)$

Integrate on both sides of equation $\left(II\right)$ and calculate the solution of the given differential equation.

$$\begin{gathered} \int \frac{dy}{y}=\int \frac{dx}{x} \\ \log \left(y\right)=\log \left(x\right)+\log \left(c\right) \\ \log \left(y\right)=\log \left(cx\right) \\ y=cx \end{gathered}$$

Hence, the solution of the given differential equation is $y=cx$.

Given: \(\left(x^4 d y+4 x^3 y d x\right)=-2 \sin x d x\)

\(\Rightarrow \int d\left(x^4 y\right)=\int-2 \sin x d x\)

\(\Rightarrow x^4 y=2 \cos x+c\)

As \(y\left(\frac{\pi}{2}\right)=0\)

So, $c=0$

Now, \(\left(\frac{\pi}{3}\right)^4 y\left(\frac{\pi}{3}\right)=2 \cos \frac{\pi}{3}\)

\(\Rightarrow \pi^4 y\left(\frac{\pi}{3}\right)=81\)

\(\left(1-x^2\right) \frac{d y}{d x}+x y=a x\)

\(\Rightarrow \frac{d y}{d x}+\frac{x}{1-x^2} y=\frac{a x}{1-x^2}\)

\(P(x)=\frac{x}{1-x^2}\)

\(\int P(x) d x=\int \frac{x}{1-x^2} d x\)

Let \(u=1-x^2, d u=-2 x d x\)

\(\int \frac{x}{1-x^2} d x=-\frac{1}{2} \int \frac{d u}{u}=-\frac{1}{2} \ln \left(1-x^2\right)\)

I.F. \(=e^{\int P d x}=e^{-\frac{1}{2} \ln \left(1-x^2\right)}=\left(1-x^2\right)^{-1 / 2}\)

For a first-order differential equation \(\frac{d y}{d x}=F(x, y)\) to be homogeneous if

\(F(t x, t y)=F(x, y) \forall t>0\)

(A) \(F(x, y)=\cos x-\sin \left(\frac{y}{x}\right)\)

\(\Rightarrow F(t x, t y)=\cos (t x)-\sin \left(\frac{y}{x}\right) \neq F(x, y)\) (not a homogeneous differential equation)

(B) \(F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}=\frac{y}{x}=F(x, y)\) (homogeneous differential equation)
(C) \(F(\lambda x, \lambda y)=\frac{(\lambda x)^2+\left(\lambda y^2\right)}{\lambda x \lambda y}=\frac{\lambda^2 x^2+\lambda^2 y^2}{\lambda^2 x y}=\frac{x^2+y^2}{x y}=F(x, y)\) (homogeneous differential equation)
(D) \(F(\lambda x, \lambda y)=\cos ^2\left(\frac{\lambda y}{\lambda x}\right)=\cos ^2\left(\frac{y}{x}\right)=F(x, y)\) (homogeneous differential equation)

$6{\int }_1^{x}f\left(t\right)dt=3xf\left(x\right)+x^3-4$

Differentiating w.r.t. \(x\)

$6f\left(x\right)=3x{f}^{\text{'}}\left(x\right)+3f\left(x\right)+3x^2$

$f\left(x\right)=x{f}^{\text{'}}\left(x\right)+x^2$

Replacing $y=f\left(x\right)$ and $\frac{dy}{dx}=f^{\text{'}}\left(x\right)$

$x\frac{dy}{dx}-y=-x^2$

$\Rightarrow \frac{x\frac{dy}{dx}-y}{x^2}=-1$

Using exact differential equation :

$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right)=-1$

$\Rightarrow \frac{y}{x}=-x+C$

$\Rightarrow f\left(x\right)=-x^2+Cx$

at $x=1,y=1\Rightarrow C=2$

$f\left(x\right)=-x^2+2x$

Now,

$f\left(2\right)-f\left(3\right)=3$

Degree is defined only when the DE is a polynomial in \(y^{\prime}, y^{\prime \prime}, \ldots\)
Here \(\left(y^{\prime \prime}\right)^2+\left(y^{\prime}\right)^3=x \sin \left(y^{\prime}\right)\) contains \(\sin \left(y^{\prime}\right)\)
⇒ Not a polynomial in derivatives because the first order derivative is inside the trigonometric function.
⇒ Degree not defined

As observing the given differential equation

$xy\left(\frac{d^{2}y}{d{x}^2}\right)+x {\left(\frac{dy}{dx}\right)}^2-y\left(\frac{dy}{dx}\right)=0$,

the highest derivative of the differential equation is $\frac{d^{2}y}{d{x}^2}$

has power $2$ so the order of the given differential equation is $2.$

The power of the term $\frac{d^{2}y}{d{x}^2}$ in the given differential equation

is $1$ so the degree of the given differential equation is $1.$

The given differential equation is:

$\frac{dy}{dx}+2y=e^{3x}......\left(I\right)$

Compare equation $\left(I\right)$ with the equation $\frac{dy}{dx}+Py=Q$ and calculate the value of $P$.

The value of $P$ is $P=2.$

Use the formula for integrating factor which is $\mu =e^{\int Pdx}$.

Put the value of $P$ in $\mu$ and calculate the integrating factor of the differential equation.

$\mu =e^{\int 2dx}=e^{2\int dx}=e^{2x}.$

Hence, the integrating factor of the given differential equation is $\mu =e^{2x}$.

$\frac{xdy-ydx}{x^2}=\frac{\sqrt{x^2+y^2}}{x^2}dx$

$d\left(\frac{y}{x}\right)=\sqrt{1+\frac{y^2}{x^2}}\cdot \frac{1}{x}dx$

$\int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}}=\int \frac{1}{x}dx$

$\Rightarrow \text{ln}\left(\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right)=\text{ln\hspace{0.17em}}x+\text{ln\hspace{0.17em}}k=\text{ln\hspace{0.17em}}kx$

$\Rightarrow y+\sqrt{x^2+y^2}=k{x}^2$

$0+1=k$

$\Rightarrow y+\sqrt{x^2+y^2}=x^2$

Now,

$y\left(3\right)+\sqrt{9+{\left(y\left(3\right)\right)}^2}=9$

$y\left(3\right)=4$

The given differential equation is:

$\frac{dy}{dx}+2y=e^{3x}......\left(I\right)$

Compare equation $\left(I\right)$ with the equation $\frac{dy}{dx}+Py=Q$ and calculate the value of $P$.

The value of $P$ is $P=2.$

Use the formula for integrating factor which is $\mu =e^{\int Pdx}$.

Put the value of $P$ in $\mu$ and calculate the integrating factor of the differential equation.

$\mu =e^{\int 2dx}=e^{2\int dx}=e^{2x}.$

Hence, the integrating factor of the given differential equation is $\mu =e^{2x}$.

Given: \(\left(1-x^2\right) \frac{d y}{d x}+x y=a x\)

\(\Rightarrow \frac{d y}{d x}+\frac{x}{1-x^2} y=\frac{a x}{1-x^2}\)

Here, \(P(x)=\frac{x}{1-x^2}\) and \(Q(x) =\frac{a x}{1-x^2}\)

\(\int P(x) d x=\int \frac{x}{1-x^2} d x\)

Let \(u=1-x^2, d u=-2 x d x\)

\(\int \frac{x}{1-x^2} d x=-\frac{1}{2} \int \frac{d u}{u}=-\frac{1}{2} \ln \left(1-x^2\right)\)

I.F. \(=e^{\int P d x}=e^{-\frac{1}{2} \ln \left(1-x^2\right)}=\left(1-x^2\right)^{-1 / 2}\)

The order of a differential equation is determined by the highest order of the derivative present in the equation.

In the given equation, the highest derivative is $\frac{d^{4}y}{d{x}^4}$, thus the order is $4$.

The formula for the integrating factor of

a differential equation of the form

$\frac{dy}{dx}+Py=Q$ is $\mu =e^{\int Pdx}.$

To find the integrating factor

of the given differential equation $\frac{dy}{dx}+2y=\sin x$,

compare the given differential equation

$\frac{dy}{dx}+Py=Q$ and

find the value $P$,

then use the formula for the integrating factor.

The value of $P$ for the given

differential equation is $P=2.$

Put the value of $P$ in the formula for integrating facto

r and calculate the value of $\mu$

which represents the integrating factor of the given differential equation.

$\mu =e^{\int 2dx}=e^{2\int dx}=e^{2x}.$

Hence,

the integrating factor of the given differential equation

is $\mu =e^{2x}.$

\(\frac{d}{d x}\left(\left(y^{\prime}\right)^3\right)=0\)

\(\Rightarrow 3\left(y^{\prime}\right)^2 y^{\prime \prime}=0\)

Highest order derivative \(=y^{\prime \prime} \Rightarrow p=2\)

Power of \(y^{\prime \prime}\) is \(1 \Rightarrow q=1\)

\(\Rightarrow p-q=2-1=1\)

The formula for the integrating factor of

a differential equation of the form

$\frac{dy}{dx}+Py=Q$ is $\mu =e^{\int Pdx}.$

To find the integrating factor

of the given differential equation $\frac{dy}{dx}+2y=\sin x$,

compare the given differential equation

$\frac{dy}{dx}+Py=Q$ and

find the value $P$,

then use the formula for the integrating factor.

The value of $P$ for the given

differential equation is $P=2.$

Put the value of $P$ in the formula for integrating facto

r and calculate the value of $\mu$

which represents the integrating factor of the given differential equation.

$\mu =e^{\int 2dx}=e^{2\int dx}=e^{2x}.$

Hence,

the integrating factor of the given differential equation

is $\mu =e^{2x}.$

Equation: \(\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=\frac{d^2 y}{d x^2}\).

Highest order derivative present: \(\frac{d^2 y}{d x^2}\)

\(\Rightarrow\) order \(=2\).

The equation is polynomial in derivatives and power of \(\frac{d^2 y}{d x^2}\) is \(1\).

\(\Rightarrow\) degree \(=1\).

As observing the given differential equation

$xy\left(\frac{d^{2}y}{d{x}^2}\right)+x {\left(\frac{dy}{dx}\right)}^2-y\left(\frac{dy}{dx}\right)=0$,

the highest derivative of the differential equation is $\frac{d^{2}y}{d{x}^2}$

has power $2$ so the order of the given differential equation is $2.$

The power of the term $\frac{d^{2}y}{d{x}^2}$ in the given differential equation

is $1$ so the degree of the given differential equation is $1.$

The given differential equation is:

$\frac{dy}{dx}=\frac{y}{x}.....\left(I\right)$

Rearrange the tems of the equation $\left(I\right)$

$\frac{dy}{y}=\frac{dx}{x}.....\left(II\right)$

Integrate on both sides of equation $\left(II\right)$ and calculate the solution of the given differential equation.

$$\begin{gathered} \int \frac{dy}{y}=\int \frac{dx}{x} \\ \log \left(y\right)=\log \left(x\right)+\log \left(c\right) \\ \log \left(y\right)=\log \left(cx\right) \\ y=cx \end{gathered}$$

Hence, the solution of the given differential equation is $y=cx$.

$16\left(\sqrt{9x+x\sqrt{x}}\right)(4+\sqrt{9+\sqrt{x}})\cos y dy=(1+2\sin y) dx$

$\int \frac{16\cos ydy}{(1+2\sin y)}=\int \frac{dx}{\left(\sqrt{9x+x\sqrt{x}}\right)(4+\sqrt{9+\sqrt{x}})}$

In L.H.S., Put \( 1+2 \sin y=p \Rightarrow 2 \cos y d y=d p\)

and in R.H.S., put \( 4+\sqrt{9+\sqrt{x}}=t \)
\( 0+\frac{d x}{2 \sqrt{9+\sqrt{x}}} \times \left(\frac{1}{2 \sqrt{x}}\right)=dt \)
\( \frac{d x}{\sqrt{9x+x \sqrt{x}}}=4 dt\)

Therefore $\int \frac{8dp}{p}=\int \frac{4dt}{t}$

$$\begin{gathered} 2\log p=\log t+\log c \\ \Rightarrow \log p^2=\log tc \end{gathered}$$

\( p^2=tc \)
\( (1+2 \sin y)^2=c(4+\sqrt{9+\sqrt{x}})\)

Substituting \(\left(256, \frac{\pi}{2}\right)\)

\( (3)^2=c(4+\sqrt{9+16})\)

\(9=9c \Rightarrow c=0\)

\( (1+2 \sin y)^2=(4+\sqrt{9+\sqrt{x}})\)

Substituting \((49, \alpha)\)

\( (1+2 \sin \alpha)^2=(4+\sqrt{9+\sqrt{49}})\)

\( (1+2 \sin \alpha)^2=(4+4)\)

\( 2 \sin \alpha+1=2 \sqrt{2} \)
\( 2 \sin \alpha=2 \sqrt{2}-1\)

The order of a differential equation is determined by the highest order of the derivative present in the equation.

In the given equation, the highest derivative is $\frac{d^{4}y}{d{x}^4}$, thus the order is $4$.

For a first-order DE \(\frac{d y}{d x}=F(x, y)\) to be homogeneous:

\(F(t x, t y)=F(x, y) \forall t>0\)

\(F(x, y)=\phi\left(\frac{y}{x}\right)\)

Option A: \(F(x, y)=\cos x-\sin \left(\frac{y}{x}\right)\)

\(\Rightarrow F(t x, t y)=\cos (t x)-\sin \left(\frac{y}{x}\right) \neq F(x, y)\) (depends on \(\left.x\right)\)

Options B, C, D depend only on \(\frac{y}{x}\) (homogeneous)

Equation: \(\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=\frac{d^2 y}{d x^2}\).

Highest order derivative present: \(\frac{d^2 y}{d x^2}\)

\(\Rightarrow\) order \(=2\).

The equation is polynomial in derivatives and \(\frac{d^2 y}{d x^2}\) occurs to power \(1\)

\(\Rightarrow\) degree \(=1\).

Degree is defined only when the DE is a polynomial in \(y^{\prime}, y^{\prime \prime}, \ldots\)
Here \(\left(y^{\prime \prime}\right)^2+\left(y^{\prime}\right)^3=x \sin \left(y^{\prime}\right)\) contains \(\sin \left(y^{\prime}\right)\)
⇒ Not a polynomial in derivatives
⇒ Degree not defined

\(\frac{d}{d x}\left(\left(y^{\prime}\right)^3\right)=0\)

\(\Rightarrow 3\left(y^{\prime}\right)^2 y^{\prime \prime}=0\)

Highest derivative \(=y^{\prime \prime} \Rightarrow p=2\)

Power of \(y^{\prime \prime}\) is \(1 \Rightarrow q=1\)

\(\Rightarrow p-q=2-1=1\)