Introduction to Differential Equations

An equation which involves derivatives of a dependent variable with respect to other independent variable is called a differential equation. It is a tool which helps in building mathematical models. Differential equations are not only used in the field of mathematics but also play a major role in other fields such as medical, chemistry, physics and engineering. In this section, we will study differential equations in detail along with solved examples.

What are Differential Equations?

An equation involving a function and one or more of its derivatives. If a function has only one independent variable then it is an ordinary differential equation.

Examples of Differential Equations:

  1. d2ydx2dydx4x=0\frac{{{d}^{2}}y}{d{{x}^{2}}}-\frac{dy}{dx}-4x=0
  2. (dy2dx2)6dydx+8y=0\left( \frac{d{{y}^{2}}}{d{{x}^{2}}} \right)-6\frac{dy}{dx}+8y=0
  3. [2+(dydx)]=(d2ydx2)23\left[ 2+\left( \frac{dy}{dx} \right) \right]={{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{\frac{2}{3}}}

Differential equation involving a function of several variables of its partial derivatives is called partial differential equation.

Order and degree of Differential equations

  • Order: Highest differential coefficient
  • Degree: Differential equation when expressed as [Polynomial in the derivatives] is the power of highest order derivative
f(xy)dmydxm+g(xm)dm1ydxm1+=0f\left( xy \right)\frac{{{d}^{m}}y}{d{{x}^{m}}}+g\left( xm \right)\frac{{{d}^{m-1}}y}{d{{x}^{m-1}}}+…=0

Note:

Degree of differential equation is only define where it is in polynomial form.

Degree may not be defined at ways:

  • y+6y2+y=0y’+6{{y}^{2}}+y=0 at (1,1)
  • (y)2+y21=0at (1,2){{\left( y’ \right)}^{2}}+{{y}^{2}}-1=0\,\,\,at\ \left( 1,2 \right)
  • y+(2y1)6+siny=0y”+{{\left( 2{{y}^{1}} \right)}^{6}}+\sin y=0 at (2,1)
  • y+sin(y)+y=0at (2,ND)y”+\sin \left( y’ \right)+y=0\,\,at\ \left( 2,ND \right)

Formation of Differential Equation

Let the y, and x be the dependent and independent variable respectively for equation  where k is arbitrary constant. Now, step to form differential equation:

 Count the number of independent arbitrary constant, (let if ‘n’)

Number of arbitrary constants will be equal to the order of differential equation. (n)

Differentiate the given equation ’n’ times to eliminate arbitrary constant.

Above elimination equation will be required differential equation.

For family of curves

f(x,y,α,α2.αn)=0f\left( x,y,\alpha ,{{\alpha }_{2}}….{{\alpha }_{n}} \right)=0 … (1)

Where α,α2αn\alpha ,{{\alpha }_{2}}…{{\alpha }_{n}} are n different parameters. If we differentiate n times, we get differential equation of the given equation (1).

e.g. Q. Form differential equation of all lines passing through origin.

Answer:

y=mxy=mx

dydx=m\frac{dy}{dx}=m

∴ xdy – ydx = 0 is answer.

Solution of Differential Equations

The solution of differential equation is the Relation between the variables involved which satisfies differential equation.

Types of solutions:

1. General solution:

It contains as many as arbitrary constants as the order of the differential equation.

2. Particular solution

The solution obtained by giving particular values to the arbitrary constants in the general solution of differential equation.

5 methods for solving the differential equation

  1. Solution by inspection
  2. Variable separable
  3. Homogeneous
  4. Linear differential equation
  5. General

Solution by Inspection

If the differential equation is of the form f(f1(x,y))d(f1(x,y))+φ(f2(x,y))d(f2(x,y))+=0f({{f}_{1}}(x,\,y))d({{f}_{1}}(x,\,y))+\varphi ({{f}_{2}}(x,\,y))d({{f}_{2}}(x,\,y))+……=0, then each term can be separately integrated.
The solution to a differential equation can be found using the inspection method. It is accompanied by memorising the following results.

(i)\  d(x+y)=dx+dy\\ (ii)\ d(xy)=xdy+ydx\\ (iii) \ d\left( \frac{x}{y} \right)=\frac{ydx-xdy}{{{y}^{2}}} \\ (iv) \ d\left( \frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}}\\ (v) \ d\,\left( \frac{{{x}^{2}}}{y} \right)=\frac{2xydx-{{x}^{2}}dy}{{{y}^{2}}}\\(vi)\ d\left( \frac{{{y}^{2}}}{x} \right)=\frac{2xydy-{{y}^{2}}dx}{{{x}^{2}}}\\ (vii)\ d\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)=\frac{2x{{y}^{2}}dx-2{{x}^{2}}ydy}{{{y}^{4}}}\\ (viii)\ d\,\left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=\frac{2{{x}^{2}}ydy-2x{{y}^{2}}dx}{{{x}^{4}}}\\ (ix)\ d\,\left( {{\tan }^{-1}}\frac{x}{y} \right)=\frac{ydx-xdy}{{{x}^{2}}+{{y}^{2}}}\\ (x)\ d\left( {{\tan }^{-1}}\frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}\\ (xi) d[ln(xy)]=xdy+ydxxy(xii) d(ln(xy))=ydxxdyxy(xiii) d[12ln(x2+y2)]=xdx+ydyx2+y2(xiv) d[ln(yx)]=xdyydxxy(xv) d (1xy)=xdy+ydxx2y2(xvi) d (exy)=yexdxexdyy2(xvii) d(eyx)=xeydyeydxx2(xviii) d(xmyn)=xm1yn1(mydx+nxdy)(xix) d (x2+y2)=xdx+ydyx2+y2(xx) d (12logx+yxy)=xdyydxx2y2(xxi) d[f(x,y)]1n1n=f(x,y)(f(x,y))n(xi)\ d[\ln (xy)]=\frac{xdy+ydx}{xy}\\ (xii)\ d\left( \ln \left( \frac{x}{y} \right) \right)=\frac{ydx-xdy}{xy}\\ (xiii)\ d\,\left[ \frac{1}{2}\ln ({{x}^{2}}+{{y}^{2}}) \right]=\frac{xdx+ydy}{{{x}^{2}}+{{y}^{2}}}\\ (xiv)\ d\left[ \ln \left( \frac{y}{x} \right) \right]=\frac{xdy-ydx}{xy}\\ (xv)\ d\text{ }\left( -\frac{1}{xy} \right)=\frac{xdy+ydx}{{{x}^{2}}{{y}^{2}}}\\(xvi)\ d\text{ }\left( \frac{{{e}^{x}}}{y} \right)=\frac{y{{e}^{x}}dx-{{e}^{x}}dy}{{{y}^{2}}}\\ (xvii)\ d\left( \frac{{{e}^{y}}}{x} \right)=\frac{x{{e}^{y}}dy-{{e}^{y}}dx}{{{x}^{2}}}\\ (xviii)\ d({{x}^{m}}{{y}^{n}})={{x}^{m-1}}{{y}^{n-1}}(mydx+nxdy)\\ (xix)\ d\text{ }\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)=\frac{xdx+y\,dy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}\\ (xx)\ d\text{ }\left( \frac{1}{2}\log \frac{x+y}{x-y} \right)=\frac{x\,dy-y\,dx}{{{x}^{2}}-{{y}^{2}}}\\ (xxi)\ \frac{d{{[f(x,\,y)]}^{1-n}}}{1-n}=\frac{{f}'(x,\,y)}{{{(f(x,\,y))}^{n}}}

Variable separable Method

If an equation can be written such that variables are separated for integration then the equation can be solved.

We get f(x)dx+g(y)dy=c\int{f\left( x \right)dx+\int{g\left( y \right)dy=c}}

Illustration

Solve log(dydx)=4x2y2\log \left( \frac{dy}{dx} \right)=4x-2y-2 given y = 1 when x = 1

Answer:

dydx=e4x2y2\frac{dy}{dx}={{e}^{4x-2y-2}} given

e2y+2dy=e4xdx\int{{{e}^{2y+2}}dy=\int{{{e}^{4x}}dx}} e2y+22=e4x4+c\frac{{{e}^{2y+2}}}{2}=\frac{{{e}^{4x}}}{4}+c put x = 1 y = 1

e42=e44+core44\frac{{{e}^{4}}}{2}=\frac{{{e}^{4}}}{4}+c\,\,or\,\,\frac{{{e}^{4}}}{4} 2e2y+2=e4x+e42{{e}^{2y+2}}={{e}^{4x}}+{{e}^{4}}

Homogenous Differential Equation

A differential equation of form dydx=f(x,y)ϕ(x,y)\frac{dy}{dx}=\frac{f\left( x,y \right)}{\phi \left( x,y \right)} where f&ϕf\And \phi are homogeneous is called as homogenous.

Homogenous function: The function (x, y) is called functions. So, if

f(λx,λy)=λnf(x,y)f\left( \lambda x,\lambda y \right)={{\lambda }^{n}}f\left( x,y \right)

Thus homogenous function can be written as f(x,y)=xnf(yx)f\left( x,y \right)={{x}^{n}}f\left( \frac{y}{x} \right) or f(x,y)=ynf(yx)f\left( x,y \right)={{y}^{n}}f\left( \frac{y}{x} \right)

Homogeneous differential equations

In first order first degree differential equation is expressed in form.

dydx=f(x,y)g(x,y)\frac{dy}{dx}=\frac{f\left( x,y \right)}{g\left( x,y \right)}

Example: Solve differential equation x2dy+y(x+y)dx=0{{x}^{2}}dy+y\left( x+y \right)dx=0 y = 1 when x = 1.

Solution:

x2+dy+y(x+y)dx=0{{x}^{2}}+dy+y\left( x+y \right)dx=0 x2+dy=y(x+y)dx{{x}^{2}}+dy=-y\left( x+y \right)dx dydx=y(x+y)x2\frac{dy}{dx}=\frac{-y\left( x+y \right)}{{{x}^{2}}}

Since each xy + y2 and x2 is homogeneous putting

y=vxy=vx dydx=v+xdvdx\frac{dy}{dx}=v+x\frac{dv}{dx} v+xdvdx=(vx2+v2x2x2)v+x\frac{dv}{dx}=-\left( \frac{v{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{{{x}^{2}}} \right) v+xdvdx=(v+v2)v+x\frac{dv}{dx}=-\left( v+{{v}^{2}} \right) 1v2+2vdv=1x\int{\frac{1}{{{v}^{2}}+2v}dv=-\int{\frac{1}{x}}} 12logv+11v+1+1=logx+logc\frac{1}{2}\log \left| \frac{v+1-1}{v+1+1} \right|=-\log x+\log c logvv+2+2logx=2logc\log \left| \frac{v}{v+2} \right|+2\log x=2\log c logvx2v+2=2logk\log \left| \frac{v{{x}^{2}}}{v+2} \right|=2\log k k=vx2v+2k=\left| \frac{v{{x}^{2}}}{v+2} \right| k=x2yy+2xk=\left| \frac{{{x}^{2}}y}{y+2x} \right|

Put y = 1 & x = 1

k=13k=\frac{1}{3} 3x2=y+2x3{{x}^{2}}=y+2x y=2x3x21y=\frac{2x}{3{{x}^{2}}-1} which is a required solution.

 

Example: Solve x2ydx(x3+y3)dy=0{{x}^{2}}ydx-\left( {{x}^{3}}+{{y}^{3}} \right)dy=0

Solution:

dydx=x2yx3+y3\frac{dy}{dx}=\frac{{{x}^{2}}y}{{{x}^{3}}+{{y}^{3}}}

Putting y = vx and dydx=v+xdvdx\frac{dy}{dx}=v+x\frac{dv}{dx} v+xdvdx=vx3x3+v3x3v+x\frac{dv}{dx}=\frac{v{{x}^{3}}}{{{x}^{3}}+{{v}^{3}}{{x}^{3}}}

= vx3x3(1+v3)\frac{v{x}^{3}}{{x}^{3}\left( 1+{v}^{3} \right)}

= v(1+v3)\frac{v}{\left( 1+{{v}^{3}} \right)} xdvdx=v1+v3vx\frac{dv}{dx}=\frac{v}{1+{{v}^{3}}}-v x(1+v3)dv=v4dxx\left( 1+{{v}^{3}} \right)dv=-{{v}^{4}}dx (1+v3v4)dv=dxx\left( \frac{1+{{v}^{3}}}{{{v}^{4}}} \right)dv=\frac{-dx}{x} 13v3+logv+logx=c-\frac{1}{3{{v}^{3}}}+\log \left| v \right|+\log \left| x \right|=c 13x3y3+logyx.x=c-\frac{1}{3}\frac{{{x}^{3}}}{{{y}^{3}}}+\log \left| \frac{y}{x}.\,x \right|=c x33y3+logy=c\frac{-{{x}^{3}}}{3{{y}^{3}}}+\log \left| y \right|=c is the required solution.

We solve by y=txy=tx & proceeding accordingly.

 

Example: xsin(yx)dy=(ysin(yx)x)dxx\sin \left( \frac{y}{x} \right)dy=\left( y\sin \left( \frac{y}{x} \right)-x \right)dx

Answer:

Put y=txy=tx

sint(t+xdtdm)=tsint1\,\,\sin t\left( t+\frac{xdt}{dm} \right)=t\sin t-1 sintxdtdx=1\sin t\frac{xdt}{dx}=-1 sintdt=dxx\int{\sin t\,dt}=-\int{\frac{dx}{x}} cost=logex+c\cos t={{\log }_{e}}x+c cos(yx)=logex+c\cos \left( \frac{y}{x} \right)={{\log }_{e}}x+c

Linear Differential Equation

Equation of form dydx+py=θ\frac{dy}{dx}+py=\theta.

Linear differential equation: A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree.

The general form of linear differential equation of first order is dydx+Py=Q\frac{dy}{dx}+Py=Q, P, Q are constants.

We solve such type of equation by multiplying both sides ePdx{{e}^{\int{Pdx}}} ePdx(dydx+Py)=QePdx{{e}^{\int{Pdx}}}\left( \frac{dy}{dx}+Py \right)=Q\,{{e}^{\int{Pdx}}} ddx{yePdx}=QePdx\frac{d}{dx}\left\{ y\,{{e}^{\int{Pdx}}} \right\}=Q\,{{e}^{\int{Pdx}}}

Integrating both sides

yePdx=QePdxdx+cy\,{{e}^{\int{Pdx}}}=\int{\,Q\,\,{{e}^{\int{Pdx}}}dx+c}

Here,

ePdx{{e}^{\int{Pdx}}} is called integrating factor.

y[I.F]=Q(I.F)dx+cy[I.F]=\int{Q\,(I.F)dx+c}

 

solve by multiplying integrating function

If epdx{{e}^{\int{pdx}}}

To get final solution of

y(if)=Q(IF)+cy(if)=\int{Q(IF)+c}

 

Illustration: Solve x2dydx+y+1=1{{x}^{2}}\frac{dy}{dx+y+1}=1

Ans : Given equation is

dydx+1x2y=1xx2\frac{dy}{dx}+\frac{1}{{{x}^{2}}}y=\frac{1x}{{{x}^{2}}}which is linear

p=1x2,q=1x2p=\frac{1}{{{x}^{2}}},q=\frac{1}{{{x}^{2}}}

IF=e1xxdx=e1x={{e}^{\int{\frac{1}{{{x}^{x}}}dx}}}=e\frac{1}{x}

Ans:

ye1x=p1x2(1x2)dx+cy\,e\,\frac{-1}{x}=\int{{{p}^{\frac{-1}{{{x}^{2}}}}}}(\frac{1}{{{x}^{2}}})dx+c y=1+ce1xy=1+c{{e}^{\frac{1}{x}}}

Applications of Differential Equations

Illustrations:

Illustration 1: If population of a bacteria get doubles in 10 years, in how many year will it triple. Given that rate of growth is proportional to the number of bacteria.

Solution: Let the total number of bacteria be ‘x’

dxdtx\frac{dx}{dt}\propto x

dxdt=kx\frac{dx}{dt}=kx

dxx=kdt\int{\frac{dx}{x}}=\int{k\,\,dt}

logx=kt+C\log x=kt+C

x=ekt+Cx=x0ektx={{e}^{kt+C}}\Rightarrow \,\,x={{x}_{0}}{{e}^{kt}}

Where (eC=x0)\left( {{e}^{C}}={{x}_{0}} \right) is population at t = 0 at t = 10 year, x=2x0x=2{{x}_{0}}

So,2x0=x0ek×10So,\,2{{x}_{0}}={{x}_{0}}\,\,{{e}^{k\times 10}}

k=log210k=\frac{\log 2}{10}

Now, for x=3x0x=3{{x}_{0}}

t=?t=?

3x0=x0.e(log210)tt=10log3log23{{x}_{0}}={{x}_{0}}.\,\,{{e}^{\left( \frac{\log 2}{10} \right)t}}\Rightarrow t=10\frac{\log 3}{\log 2}

t=15.9t=15.9 years

1st order differential equations

dNdt=kN\frac{dN}{dt}=kN

 Illustration 2: Biologist, at t = 0 puts 100 bacteria, in favourable growth medium, 6 hrs later it became 450. Assume exponential growth find growth const k.

Solution: A=PektA=P{{e}^{kt}} 450=100e6k450=100{{e}^{6k}}

k=n4.56k=\frac{\ell n\,4.5}{6}

Population and Number of People

Illustration 3:  If after 2 years population has doubled & after 3 years the population is 20k, find the no of people initially living.

Solution: k=12n2k=\frac{1}{2}\ell n\,2 No=20k22=7071No=\frac{20k}{2\sqrt{2}}=7071

Illustration 4:  5 mice in stable population of 500 are infected with disease test a theory which says rate of change of infected population is α product of diseased one & disease free.

How long it takes for half the population to contract the disease?

Solution:

dNdt=kN(500N)\frac{dN}{dt}=k\,\,\,N\left( 500-N \right) n(N500N)=500(kt+c)\ell n\left( \frac{N}{500-N} \right)=500\left( kt+c \right) N500N=c1e500ktt=0N=5\frac{N}{500-N}={{c}_{1}}\,\,{{e}^{500kt}}\,\,\,t=0\,\,\,N=5 5495=c1e500k(0)c1=199\frac{5}{495}={{c}_{1}}{{e}^{500k\left( 0 \right)}}\Rightarrow {{c}_{1}}=\frac{1}{99}

N500N=199e500kt\frac{N}{500-N}=\frac{1}{99}\,{{e}^{500kt}} N=2501=199e500kT1/2N=250\Rightarrow 1=\frac{1}{99}{{e}^{500k{{T}_{1/2}}}} T1/2=n(99)500k=0.0091k{{T}_{1/2}}=\frac{\ell n\left( 99 \right)}{500k}=\frac{0.0091}{k} time units.

Newton’s cooling law

Time rate of change of temp of a body is α to temp difference between body & its surrounding medium

dTdt=k(TTm)\frac{dT}{dt}=-k\left( T-{{T}_{m}} \right) Tm{{T}_{m}}\to Surrounding temp

Cooling produces dTdt\frac{dT}{dt}\to -ve T greater than Tm.

Illustration 5: A metal bar at 100°F at room temp 0°F if after 20 min, temp of bar is 50°F.

(a) Time to reach 25°F

(b) Temp after 10 min.

Solution: dTdt=k(TTm)dTdt+kT=0\frac{dT}{dt}=-k\left( T-{{T}_{m}} \right)\Rightarrow \frac{dT}{dt}+kT=0 Tm=0{{T}_{m}}=0 T=cektT=c{{e}^{-kt}}

T = 100 at t = 0

c=100ek0c=100\,{{e}^{k0}} T=100ektT=100\,{{e}^{-kt}}

At t = 20 T = 50

12=ek20n220=k\frac{1}{2}={{e}^{-k\,20}}\,\,\,\,\,\frac{\ell n2}{20}=k

Now,

(a) 25=100en220t25=100\,\,{{e}^{-\frac{\ell n2}{20}t}} n4=n220t\ell n\,4=\frac{\ell {{n}^{2}}}{20}t

or t = 40

(b) T=100en220×10=1002=70.71FT=100{{e}^{-\frac{\ell n2}{20}\times 10}}=\frac{100}{\sqrt{2}}=70.71{}^\circ F

First Order Differential Equations

Below are some of the most important and popular methods to find the solution of first order and first degree differential equations along with the examples.

Methods of solving

1. Differential equation is form of dydx=f(x)\frac{dy}{dx}=f\left( x \right). In this we integrate both sides to get general solutions.

Example:

dydx=xx2+1\frac{dy}{dx}=\frac{x}{{{x}^{2}}+1}

Solution:

dy=(xx2+1)dxdy=\left( \frac{x}{{{x}^{2}}+1} \right)dx dy=xdxx2+1\int{dy}=\int{\frac{xdx}{{{x}^{2}}+1}} dy=122xdxx2+1\int{dy}=\frac{1}{2}\int{\frac{2x\,\,dx}{{{x}^{2}}+1}} y=12logx2+1+c;xεRy=\frac{1}{2}\log \left| {{x}^{2}}+1 \right|+c;\,\,\,\,\,\,\,\,x\varepsilon R is the solution of given differential equation

Example:

dydx=3e2x+3e4xex+ex\frac{dy}{dx}=\frac{3{{e}^{2x}}+3{{e}^{4x}}}{{{e}^{x}}+{{e}^{-x}}}

Solution:

dydx=3e2x+3r4xex+1ex=3e2x(1+e2x)ex+1ex\frac{dy}{dx}=\frac{3{{e}^{2x}}+3{{r}^{4x}}}{{{e}^{x}}+\frac{1}{{{e}^{x}}}}=\frac{3{{e}^{2x}}\left( 1+{{e}^{2x}} \right)}{{{e}^{x}}+\frac{1}{{{e}^{x}}}} dydx=3e3x\frac{dy}{dx}=3{{e}^{3x}} dy=3e3xdx\int{dy}=\int{3{{e}^{3x}}dx} y=3(e3x3)+cy=3\left( \frac{{{e}^{3x}}}{3} \right)+c is required solutions.

2. Differential equations reducible to variable separable type.

When first order differential equation could not solved by integration, then we apply substitution method in order to reduce the variable.

A differential equation of the form dydx=f(px+qy)\frac{dy}{dx}=f\left( px+qy \right)

Substitute px+qy=tpx+qy=t

For example: Solve dydx=(ax+by)\frac{dy}{dx}=\left( ax+by \right)

Solution:

Assume ax+by=tax+by=t

a+bdydx=dtdxa+b\frac{dy}{dx}=\frac{dt}{dx}

dydx=1b(dtdxa)\frac{dy}{dx}=\frac{1}{b}\left( \frac{dt}{dx}-a \right)

1b(dtdxa)=t\frac{1}{b}\left( \frac{dt}{dx}-a \right)=t

dtdx=bt+a\frac{dt}{dx}=bt+a

dx=1bbdtbt+a\int{dx}=\frac{1}{b}\int{\frac{bdt}{bt+a}}

x=1bln(bt+a)x=\frac{1}{b}\ln \left( bt+a \right)

x=1bln(b(ax+by)+a)x=\frac{1}{b}\ln \left( b\left( ax+by \right)+a \right)

3. First order differential equation of Homogeneous form:

Differential equation of the form: dydx=g(x,y)=f(y/x)\frac{dy}{dx}=g\left( x,y \right)=f\left( y/x \right)

=f(y/x)=f\left( y/x \right) are called homogeneous form.

For such equations substitutes yx=t\frac{y}{x}=t

y=xty=xt

dydx=t+xdtdx\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}

t+xdtdx=f(t)\Rightarrow t+x\frac{dt}{dx}=f\left( t \right)

xdtdx=f(t)t\Rightarrow x\frac{dt}{dx}=f\left( t \right)-t

dtf(t)t=dxx\Rightarrow \int{\frac{dt}{f\left( t \right)-t}}=\int{\frac{dx}{x}}

And then ‘t’ can be replaced with (yx)\left( \frac{y}{x} \right) to get a solution of differential equation.

Example:  Solve the differential equation

dydx=3x+yxy,\frac{dy}{dx}=\frac{3x+y}{x-y},

dydx=g(x,y)=3x+yxy\frac{dy}{dx}=g\left( x,y \right)=\frac{3x+y}{x-y}

=f(yx)=3+yx1yx=f\left( \frac{y}{x} \right)=\frac{3+\frac{y}{x}}{1-\frac{y}{x}}

yx=t\frac{y}{x}=t

y=xtf(t)=3+t1ty=xt\,\,\,\,\,\Rightarrow f\left( t \right)=\frac{3+t}{1-t}

dydx=t+xdtdx=f(t)\frac{dy}{dx}=t+x\frac{dt}{dx}=f\left( t \right)        

xdtdx=f(t)t\Rightarrow \,\,\,x\frac{dt}{dx}=f\left( t \right)-t

=3+3t1tt=\frac{3+3t}{1-t}-t

=3+tt+t21t=\frac{3+t-t+{{t}^{2}}}{1-t}

xdtdx=3+t21tx\frac{dt}{dx}=\frac{3+{{t}^{2}}}{1-t}

dxx=(1t)dt3+t2\int{\frac{dx}{x}}=\int{\frac{\left( 1-t \right)dt}{3+{{t}^{2}}}}

lnx+C1=dt3+t2122t3+t2dt\ln x+{{C}_{1}}=\int{\frac{dt}{3+{{t}^{2}}}}-\frac{1}{2}\int{\frac{2t}{3+{{t}^{2}}}}\,dt

=13tan1(t3)12ln(3+t2)+C2=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{3}} \right)-\frac{1}{2}\ln \left( 3+{{t}^{2}} \right)+{{C}_{2}}

lnx+C1=13tan1(y3x)12ln(3+y2x2)+C2\ln x+{{C}_{1}}=\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{y}{\sqrt{3}x} \right)-\frac{1}{2}\ln \left( 3+\frac{{{y}^{2}}}{{{x}^{2}}} \right)+{{C}_{2}}

Types of Differential equation

  • dydx=f(y)\frac{dy}{dx}=f\left( y \right) we integrate both the sides to get general solutions.
  • dydx=f(y)\frac{dy}{dx}=f\left( y \right)
  • dxdy=1f(y)f(y)0\frac{dx}{dy}=\frac{1}{f\left( y \right)}f\left( y \right)\ne 0
  • dx=dyd(y)dx=\frac{dy}{d\left( y \right)}
  • x=dyf(y)+cx=\int{\frac{dy}{f\left( y \right)}}+c

Equation reducible to variable separable form

Differential equation in form of:- dydx=f(ax+by+c)\frac{dy}{dx}=f\left( ax+by+c \right) can be reduced to variable separable form by substitution ax + by + c = v.

Practice Problems on Differential Equations

Example 1: Solve dydx+y=1\frac{dy}{dx}+y=1

Solution:

dydx=1y\frac{dy}{dx}=1-y dxdy=11y\frac{dx}{dy}=\frac{1}{1-y} dx=dy1y\int{dx}=\int{\frac{dy}{1-y}}

x = – log | 1 – y | + c is required solution

 

Example 2: Solve dydx=secy\frac{dy}{dx}=\sec y

Solutions:

dxdy=1secy=cosy\frac{dx}{dy}=\frac{1}{\sec y}=\cos y dx=cosydy\int{dx}=\int{\cos \,y\,dy}

x = sin y + c is required solutions.

  1. Equation is variable separable form we integrate both the sides and solutions is given by
f(x)dx=g(y)dy+c\int{f\left( x \right)\,}dx=\int{g\left( y \right)\,}dy+c

 

Example 3: Solve (x+1)dydx=2xy\left( x+1 \right)\frac{dy}{dx}=2xy

Solutions:

(x+1)dy=2xydx\left( x+1 \right)dy=2xy\,\,dx dyy=2xdxx+1\frac{dy}{y}=\frac{2x\,dx}{x+1} dyy=2xdxx+1\frac{dy}{y}=\frac{2x\,\,dx}{x+1} dyy=2xdx(x+1)\int{\frac{dy}{y}}=\int{\frac{2x\,\,dx}{\left( x+1 \right)}} logy=2{xlogx+1}+c\log \,y=2\left\{ x-\log \left| x+1 \right| \right\}+c is the solutions.

 

Example 4: Solve ex1y2dx+yxdy=0{{e}^{x}}\sqrt{1-{{y}^{2}}\,}dx+\frac{y}{x}dy=0

Solutions:

xexdx=y1y2dyx{{e}^{x}}dx=\frac{-y}{\sqrt{1-{{y}^{2}}}}dy

Apply product rule of integration

xexexdx=12dttt=1y2x{{e}^{x}}-\int{{{e}^{x}}}dx=\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}t=1-{{y}^{2}} xexex=t+cx{{e}^{x}}-{{e}^{x}}=\sqrt{t}+c xexex=1+y2+cx{{e}^{x}}-{{e}^{x}}=\sqrt{1+{{y}^{2}}}+c is the required solutions.

 

Example 5: Solve dydx=ex+y\frac{dy}{dx}={{e}^{x+y}}

Solution:

dydx=exey\frac{dy}{dx}={{e}^{x}}{{e}^{y}} dy=ex.ey.dxdy={{e}^{x}}.{{e}^{y}}.dx dyey=exdx\frac{dy}{{{e}^{y}}}={{e}^{x}}\,dx eydy=exdx\int{{{e}^{-y}}}\,dy=\int{{{e}^{x}}}dx ey1=ex+C\frac{{{e}^{-y}}}{-1}={{e}^{x}}+C ey=ex+C-{{e}^{-y}}={{e}^{x}}+C

 

Example 6: Solve dydx=cos(x+y)\frac{dy}{dx}=\cos \left( x+y \right)

Solution:

x + y = v

1+dydx=dvdx1+\frac{dy}{dx}=\frac{dv}{dx} dvdx=1\frac{dv}{dx}=-1

Put these value in given D.E

dvdx1=cosv\frac{dv}{dx}-1=\cos \,v dvdx=1+cosv\frac{dv}{dx}=1+\cos v dv1+cos  v=dx\frac{dv}{1+cos\;v}=dx

Integrating both sides

tanv2=x+c\tan \frac{v}{2}=x+c tan(x+y2)=x+c\tan \left( \frac{x+y}{2} \right)=x+c is the required solutions.

 

Example 7: Solve (x+y)2dydx=a2{{\left( x+y \right)}^{2}}\frac{dy}{dx}={{a}^{2}}

Solution:

x + y = v

Then dydx+1=dvdx\frac{dy}{dx}+1=\frac{dv}{dx} dvdx=dydx+1dvdx1=dydx\frac{dv}{dx}=\frac{dy}{dx}+1\Rightarrow \frac{dv}{dx}-1=\frac{dy}{dx}

Put these in D.E V2(dvdx1)=a2{{V}^{2}}\left( \frac{dv}{dx}-1 \right)={{a}^{2}} v2dvdxv2=a2{{v}^{2}}\frac{dv}{dx}-{{v}^{2}}={{a}^{2}} v2dvdx=a2+v2{{v}^{2}}\frac{dv}{dx}={{a}^{2}}+{{v}^{2}} (V2a2+v2)dv=dx\int{\left( \frac{{{V}^{2}}}{{{a}^{2}}+{{v}^{2}}} \right)}dv=\int{dx} vatan1(va)=x+cv-a\,{{\tan }^{-1}}\left( \frac{v}{a} \right)=x+c (x+y)atan1(x+ya)=x+C\left( x+y \right)-a{{\tan }^{-1}}\left( \frac{x+y}{a} \right)=x+C is a required solutions.

 

Example 8: Solve dydxyx=2x2\frac{dy}{dx}-\frac{y}{x}=2{{x}^{2}}

Solution:

dydx+(1x)y=2x2\frac{dy}{dx}+\left( -\frac{1}{x} \right)y=2{{x}^{2}} P=1xandQ=2x2P=\frac{-1}{x}\,\,and\,\,Q=2{{x}^{2}}

Multiply both sides by I.F

I.F =ePdx=e1xdx=elogx