Differential Equations

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a Differential Equation (D. E).

A Differential Equation (D. E) is said to be ordinary, if the differential coefficients have reference to a single independent variable only e.g. $\frac{d^{2}y}{d{x}^2}-\frac{2dy}{dx}+\cos x=0$ and it is said to be partial if there are two or more independent variables. e.g. $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0$ is a partial differential equation. We are concerned with ordinary differential equations only. 

1.0Order Of Differential Equations

The order of a differential equation (D.E) is the order of the highest differential coefficient occurring in it. 

Example:

$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\frac{\mathrm{dy}}{\mathrm{dx}}+1=0$, Order is 2 

$\frac{d^{3}y}{d{x}^3}+x=0$ ,  Order is 3.

2.0Degree of Differential Equations

The highest exponent of the highest order differential coefficient, when the differential equation is expressed as a polynomial in all the differential coefficients, is called degree of the differential equation.

Thus, the differential equation:

$a_{0,0}+{\sum }_{i=1}^{n_1}{a}_{i,1}{\left(\frac{d^{m}y}{d{x}^m}\right)}^i+{\sum }_{i=1}^{n_2}{a}_{i,2}{\left(\frac{d^{m-1}y}{d{x}^{m-1}}\right)}^i+\dots =0$ is of order m & degree $n_1.$

where $a_{i,j}$ is some expression in x, y and $a_{n_1,1}\ne 0$

Example: Find degree of $\sqrt{\frac{d^{2}y}{d{x}^2}}=3\sqrt{\frac{dy}{dx}+4}$

Solution: ${\left(\frac{d^{2}y}{d{x}^2}\right)}^3=\frac{dy}{dx}+4$

Degree = 3

3.0Solution Of Differential Equations

The solution to a differential equation is a relation between the variables that does not include derivatives but satisfies the given differential equation. In other words, it is a function from which the given differential equation can be derived.

Thus, the solution of $\frac{dy}{dx}=e^x$ could be obtained by simply integrating both sides, i.e., $y=e^x+c$  and that of, $\frac{dy}{dx}=px+qisy=\frac{p{x}^2}{2}+qx+c$, where c is an arbitrary constant.

(i) A general solution, or integral, of a differential equation is a relation between the variables that does not include derivatives. This relation contains as many arbitrary constants as the order of the differential equation.

For example, a general solution of the differential equation $\frac{d^{2}x}{d{t}^2}=-4xisx=A\cos 2t+B\sin 2t$

where A and B are the arbitrary constants.

(ii) Particular solution or integral is that solution of the differential equation which is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

For example, $x=10cos2t+5sin2t$ is a particular solution of differential equation $\frac{d^{2}x}{d{t}^2}=-4x$. 

4.0Elementary Types of First Order and First-Degree Differential Equations

Separation of Variables

Some differential equations can be solved by the method of separation of variables (or “variable separable”).

This method is only possible, if we can express the differential equation in the form 

$A(x)dx+B(y)dy=0$

where A(x) is a function of '' only and B(y) is a function of $\text{ ’ y’ }$ only.

A general solution of this is given by,

$\int A(x)dx+\int B(y)dy=c$

where $\text{ ’C’ }$ is the arbitrary constant.

Homogeneous Differential Equations

1. A function f(x, y) is said to be a homogeneous function of degree n, if the substitution 

$x=\lambda x,y=\lambda y,\lambda >0$ produces the equality  

$f_{\left(\lambda {}^x,{\lambda }^y\right)}={\lambda }^{n}f(x,y)$

The degree of homogeneity  $\text{ ’n’ }$ can be any real number. 

2. A differential equation of the form $\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$

Where f(x, y) and g(x, y) are homogeneous functions of x, y and of the same degree are said to be homogeneous. Such equations can be solved by substituting y = vx.

3. Equation Reducible to Homogeneous form

The equation of the form $\frac{dy}{dx}=\frac{a_{1}x+b_{1}y+c_1}{a_{2}x+b_{2}y+c_2}Where\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ can be reduced to homogeneous form by changing the variable x, y to u, v as x = u + h, y = v + k. 

Linear Differential Equations

A differential equation is considered linear if the dependent variable and its derivatives appear only to the first power and are not multiplied by each other.

The n^th order linear differential equation is of the form:

$a_0(x)\frac{d^{n}y}{d{x}^n}+a_1(x)\frac{d^{n-1}y}{d{x}^{n-1}}+\dots \dots ..+a_n(x)y=\phi (x),where{a}_0(x),a_1(x)\dots .a_n(x)$ are called the coefficients of the differential equation.

Note that a linear differential equation is always of the first degree. However, not every differential equation of the first degree. e.g. the differential equation $\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3+y^2=0$ is not linear, though its degree is 1.

(i) Linear Differential Equations of First Order:

The most general form of a linear differential equation of first order is $\frac{dy}{dx}+Py=Q$ , where $P_{\&}Q$

are functions of x.

To solve such an equation multiply both sides by $e^{\int Pdx}$ .

So that we get $e^{\int Pdx}\left[\frac{dy}{dx}+Py\right]=Q{e}^{\int Pdx}$ ...(i)

$\frac{d}{dx}\left(e^{\int Pdx}\cdot y\right)=Q{e}^{\int Pdx}$ ...(ii)

On integrating equation (ii), we get 

$y{e}^{\int Pdx}=\int Q{e}^{\int Pdx}dx+c$

This is the required general solution.

Also Read: First Order Differential Equation

(ii) Equation Reducible to Linear Form:

The equation of the form $\frac{dy}{dx}+Py=Q{y}^n$ , where P and Q are functions of x, is called Bernoulli’s equation.    

On dividing by $y^n$, we get $y^{-n}\frac{dy}{dx}+P{y}^{-n+1}=Q$

Let $y^{-n+1}=t,sothat(-n+1)y^{-n}\frac{dy}{dx}=\frac{dt}{dx}$

then equation becomes $\frac{dt}{dx}+P(1-n)t=Q(1-n)$

which is linear with t as a dependent variable. 

5.0Solved Examples of Differential Equations

Example 1: Find the order and degree of the following differential equation: $\frac{d^{2}y}{d{x}^2}=\sin \left(\frac{dy}{dx}\right)$

Solution: The given differential equation has order 2. Since the given differential equation is not a polynomial in the differential coefficients, the degree of the equation is not defined.

Example 2: Solve the differential equation $xy\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\left(1+x+x^2\right)$

Solution: Differential equation can be rewritten as

$xy=\left(1+y^2\right)\left(1+\frac{x}{1+x^2}\right)\Rightarrow \frac{y}{1+y^2}dy=\left(\frac{1}{x}+\frac{1}{1+x^2}\right)dx$

Integrating, we get

$\frac{1}{2}\ln \left(1+y^2\right)=\ln x+{\tan }^{-1}x+\ln c\Rightarrow \sqrt{1+y^2}=cx{e}^{{\tan }^{-1}x}$

Example 3: Solve $\frac{dy}{dx}=\frac{x-2y+1}{2x+2y+3}$

Solution:

$\frac{dy}{dx}=\frac{x-2y+1}{2x+2y+3}$

${\models }^{2xdy+2ydy+3dy}=xdx-2ydx+dx$

$\Rightarrow (2y+3)dy=(x+1)dx-2(xdy+ydx)$

On integrating, we get

$\Rightarrow \int (2y+3)dy=\int (x+1)dx-\int 2d(xy)$

Solving : $y^2+3y=\frac{x^2}{2}+x-2xy+c$

Example 4: Find the degree of homogeneity of function 

(i) $f(x,y)=x^2+y^2$

(ii) $f(x,y)=\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}\right)/(x+y)$

(iii) $f(x,y)=\sin \left(\frac{x}{y}\right)$

Solution:

(i) $f(\lambda x,\lambda y)={\lambda }^2{x}^2+{\lambda }^2{y}^2={\lambda }^2\left(x^2+y^2\right)={\lambda }^{2}f(x,y)$

degree of homogeneity ⟶ 2

(ii) $f(\lambda x,\lambda y)=\frac{{\lambda }^{3/2}{x}^{3/2}+{\lambda }^{3/2}{y}^{3/2}}{\lambda x+\lambda y}f(\lambda x,\lambda y)={\lambda }^{\frac{1}{2}}f(x,y)$

degree of homogeneity ⟶ 1/2

(iii) $f(\lambda x,\lambda y)=\sin \left(\frac{\lambda x}{\lambda y}\right)={\lambda }^{\circ }\sin \left(\frac{x}{y}\right)={\lambda }^{\circ }f(x,y)$

degree of homogeneity ⟶ 0

Example 5: The solution of the given differential equation $\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}s$- 

(A) ${\sin }^{2}y=x\sin y+\frac{x^2}{2}+c$ (B) ${\sin }^{2}y=x\sin y-x+c$

(C) ${\sin }^{2}y=x+\sin y+x^2+c$ (D) ${\sin }^{2}y=x-\sin y+\frac{x^2}{2}+c$

Ans. (A)

Solution:

Here, $\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}$

$\Rightarrow (\cos y)\frac{dy}{dx}=\frac{\sin y+x}{2\sin y-x}(put\sin y=t)$

$\Rightarrow \frac{dt}{dx}=\frac{t+x}{2t-x}$ (put t=v x)

$\frac{xdv}{dx}+v=\frac{v+1}{2v-1}$

$\therefore \frac{xdv}{dx}=\frac{v+1}{2v-1}-v=\frac{v+1-2v^2+v}{2v-1}$

or $\frac{2v-1}{-2v^2+2v+1}dv=\frac{dx}{x}$

on solving, we get

${\sin }^{2}y=x\sin y++c$.

Example 6: Solve $\frac{dy}{dx}=\frac{x+2y+3}{2x+3y+4}$

Solution:

Equation reducible to homogeneous form

Put $x=X+h,y=Y+k0$

We have $\frac{dY}{dX}=\frac{X+2Y+(h+2k+3)}{2X+3Y+(2h+3k+4)}$

To determine h and k, we write

$h+2k+3=0,2h+3k+4=0\Rightarrow h=1,k=-2$

So that $\frac{dY}{dX}=\frac{X+2Y}{2X+3Y}$

Putting $Y=VX$, we get

$V+X\frac{dV}{dX}=\frac{1+2V}{2+3V}\Rightarrow \frac{2+3V}{3V^2-1}dV=-\left(\frac{dX}{X}\right)$

$\Rightarrow \left[\frac{2+\sqrt{3}}{2(\sqrt{3}V-1)}-\frac{2-\sqrt{3}}{2(\sqrt{3}V+1)}\right]dV=-\left(\frac{dX}{X}\right)$

$\Rightarrow \frac{2+\sqrt{3}}{2\sqrt{3}}\log (\sqrt{3}V-1)-\frac{2-\sqrt{3}}{2\sqrt{3}}\log (\sqrt{3}V+1)=(-\log X+c)$

${|}^{\frac{2+\sqrt{3}}{2\sqrt{3}}}\log (\sqrt{3}Y-X)-\frac{2-\sqrt{3}}{2\sqrt{3}}\log (\sqrt{3}Y+X)=A$

where A is another

constant and

$X=x-1,Y=y+2\text{. }$

Example 7: Which of the following equations is a linear differential equation?

(A) $\frac{dy}{dx}+x{y}^2=1$ (B) $x^2+y=e^x$

(C) $\frac{dy}{dx}+3y=x{y}^2$ (D) $x\frac{dy}{dx}+y^2=\sin x$

Ans. (B)

Solution:

Clearly answer is (B)

Example 8: Solve the differential equation $\frac{dy}{dx}+\frac{y}{x}=x^2{y}^6$ .

Solution:

The provided differential equation can be written as $\frac{1}{y^6}\frac{dy}{dx}+\frac{1}{x{y}^5}=x^2$

Putting y^{-5}=v so that 

$-5y^{-6}\frac{dy}{dx}=\frac{dv}{dx}\text{ or }{y}^{-6}\frac{dy}{dx}=-\frac{1}{5}\frac{dv}{dx}$

we get

$-\frac{1}{5}\frac{dv}{dx}+\frac{1}{x}v=x^2\Rightarrow \frac{dv}{dx}-\frac{5}{x}v=-5x^2$ ...(i) 

This is the standard form of the linear differential equation having integrating factor 

I.F = $e^{\int -\frac{5}{x}dx}=e^{-5\log x}=\frac{1}{x^5}$

Multiplying both sides of (i) by I.F. and integrating w.r.t. x 

We get $\text{ v. }\frac{1}{x^5}=\int -5x^2\cdot \frac{1}{x^5}dx$

$\Rightarrow \frac{v}{x^5}=\frac{5}{2}{x}^{-2}+c$

$\Rightarrow y^{-5}{x}^{-5}=\frac{5}{2}{x}^{-2}+c$

which is the required solution.

6.0Practice Question on Differential Equations

1. Find the order and degree of differential equation ${\left[\frac{d^{2}x}{d{t}^2}\right]}^3+{\left[\frac{dx}{dt}\right]}^4-xt=0$

(A) order : 3 ;  degree : 2 (B) order : 2 ;  degree : 4

(C) order : 2 ;  degree : 3 (D) order : 4 ;  degree : 2

2. Obtain the differential equation of the family of circles $x^2+y^2+2gx+2fy+c=0$ ; where $g,f\&c$

are arbitrary constants.

(A) $\left[1+{\left(y^{\prime }\right)}^2\right]\cdot y^{\prime \prime \prime }-3y^{\prime }{\left(y^{\prime \prime }\right)}^2=0$

(B) $\left[1+{\left(y^{\prime }\right)}^2\right]\cdot y^{\prime \prime \prime }+\left(y^{\prime \prime }+2y^{\prime }\right){\left(y^{\prime \prime }\right)}^2=0$

(C) $\left[1+{\left(y^{\prime }\right)}^2\right]\cdot y^{\prime \prime \prime }+3y^{\prime }{\left(y^{\prime \prime }\right)}^2=0$

(D) $\left[1-{\left(y^{\prime }\right)}^2\right]\cdot y^{\prime \prime \prime }+3y^{\prime }{\left(y^{\prime \prime }\right)}^2=0$

3. Solution of the differential equation $\frac{dy}{dx}=\sin (x+y)+\cos (x+y)$ is

(A) $\tan \left(\frac{x+y}{2}\right)=a{e}^x-1$

(B) $\tan \left(\frac{x-y}{2}\right)=a{e}^x-1$

(C) $\tan \left(\frac{x+y}{2}\right)=a{e}^{-x}-1$

(D) $\tan \left(\frac{x+y}{2}\right)=a{e}^x+1$

4. Solution of the differential equation $\frac{xdx+ydy}{\sqrt{x^2+y^2}}=\frac{xdy-ydx}{x^2}$ is

(A) $\sqrt{x^2-y^2}=\frac{y}{x}+c$

(B) $\sqrt{x^2+y^2}=-\frac{y}{x}+c$

(C) $\sqrt{x^2+y^2}=\frac{x}{y}+c$

(D) $\sqrt{x^2+y^2}=\frac{y}{x}+c$

Answer Sheet: