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Category : 12th Class

                                                                               Differential Equations

• Definition: An equation involving independent variable, dependent variable and its derivatives is said to be differential equation.

e.g. An equation of the form of

\[\therefore \,\,\,\,\,y=f(x,y,p)\]is said to be differential equation, where\[p=\frac{dy}{dx}\]e.g.

(1) \[y={{\left( \frac{dy}{dx} \right)}^{3}}+5\]              (2) \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+y\frac{dy}{dx}=5y+c\] etc.

• Order and Degree of differential Equation:

The order of highest order derivative occurred in differential equation is said to be the order of the differential equation whereas power/exponent of the highest order derivative term in the different equation whereas power/exponent of the highest order derivative term in the differential equation is said to be the degree of the differential equation.

e.g.

(1) \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+2{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+6{{\left( \frac{dy}{dx} \right)}^{6}}+7y=0\]

Here order of differential equation be 3

Degree of differential equation = 1

(2) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{\frac{3}{2}}}=\frac{{{d}^{3}}y}{d{{x}^{3}}}\]

Order of this differential equation = 3

Degree of this differential equation = 2

Types of Differential Equation:

1. Ordinary Differential Equation: A differential equation involving single independent variable, is said to be an ordinary differential equation.
2. g. (1) \[\frac{dy}{dx}+6y=6{{x}^{2}}\] (2) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+2\frac{dy}{dx}+6y=0\]
3. Partial Differential Equation: A differential equation having two or more than two independent variables is said to be a partial differential equation.

If \[u=f(x,y,z),\] then its partial differential equation (P.D.E) will be

\[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{z}^{2}}}=0\]

• Methods for solving First Order, First Degree Differential equations:

An equation of the form \[\frac{dy}{dx}=\frac{M}{N},\]where M and N be the functions of x and y or any constant value, is said to be the first order, first degree differential equation.

This type of differential equation consists of following types of function

1. Separable function
2. Recudible seperable function
3. Homogeneous function and Non-Homogeneous function
4. Linear Differential equation
5. Recudible linear differential equation
6. Exact differential equation

Solved Example

1. Find the general solution of differential equation \[\frac{dy}{dx}=(1+{{x}^{2}})(1+{{y}^{2}}).\]

Solution: \[\frac{dy}{dx}=(1+{{x}^{2}}).(1+{{y}^{2}})\]

\[\Rightarrow \]   \[\frac{dy}{1+{{y}^{2}}}=(1+{{x}^{2}})dx\]

             Integrating both sides, we get

            \[{{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{2}+c\]

\[\Rightarrow \] \[y=\tan \left( \frac{2x+{{x}^{3}}+2c}{2} \right)\]

• Method for solving the Homogeneous differential equation

Step 1: First of all arrange the given equation as \[\frac{dy}{dx}=\frac{\phi (x)}{f(x)},\] where \[\phi (x)\] and \[f(x)\] having same degree.

Step2: Put\[y=\text{v}x\]and differentiate it on both sides to get \[\frac{dy}{dx}=\text{v}+x\frac{d\text{v}}{dx}\]

Step 3: Then put these value in the given equation and further hence using variable seperable method, solution will be obtained in the form of v.

Step 4: Then replace v by \[\frac{y}{x}\] & hence find the result.

Solved Example

1. Solve the differential equation \[y'=\frac{x+y}{x}.\]

Solution:\[y'=\frac{dx}{dx}=\frac{x+y}{x}.\]             (1)

It is a homogeneous equation.

Now, put \[y=\text{v}x\]                         (2)

On differentiating both sides of equ (2), we get

\[\frac{dy}{dx}=\text{v}+x\frac{d\text{v}}{dx}\]

On replacing the value of \[\frac{dy}{dx}\] in equ (1), it becomes

\[\text{v}+x\frac{d\text{v}}{dx}=\frac{x+\text{v}x}{x}=(1+\text{v})\]

\[\Rightarrow \,\,\,x\frac{d.\text{v}}{dx}=1+\text{v}\,-\,\text{v=1}\]                     \[\Rightarrow \]            \[d\text{v}=\frac{dx}{x}\]

On integrating both sides, we get

\[\text{v}=\log x+\log c\]

\[\frac{y}{x}=\log xc\](where log \[c=\] integration constant) \[y=x\log xc\]

which is the required solution.

• Linear Differential Equation: An equation of the form of \[\frac{dy}{dx}+P.y=Q,\] where P and Q is in the form of x or any constant (but not of y), is said to be a linear differential equation.

• Method of Solving Linear Differential Equation (L.D.E.)

Step 1: First of all, find the integrating factor (I.F.)

I.F. \[={{e}^{\int P.dx}}\]

Step 2: Then, the general solution will be given as \[y(I.F.)\]\[=\int{Q.(I.F.)dx}\]

Solved Example

1. Solve \[\frac{dy}{dx}+\frac{y}{x}={{x}^{3}}.\]

Solution: Given \[\frac{dy}{dx}+\left( \frac{1}{x} \right).y={{x}^{3}}\]

which is a Linear Differential Equation (L.D.E.)

Here, \[P=\frac{1}{x}\]and\[Q={{x}^{3}}\]

Now I.F\[={{e}^{\int P.dx}}\]\[={{e}^{\int \frac{1}{x}.dx}}\]\[={{e}^{log\,x}}\]\[=x\]

\[\therefore \] Hence the general solution will be

\[y.(I.F)=\int{Q.(I.F).dx}\]

\[y.x=\int{{{x}^{3}}}.x.dx=\int{{{x}^{4}}.dx}\]

\[\Rightarrow yx=\frac{{{x}^{5}}}{5}+c\]

\[\Rightarrow 5xy={{x}^{5}}+K\]

which is the required solution.

• Linear Differential Equation in the form of \[\frac{dx}{dy}\]: Sometimes, we see that the given linear differential equation is not in the form of\[\left( \frac{dy}{dx} \right)\]. But it is linear in \[\left( \frac{dy}{dx} \right)\]. After the certain manipulations in such case the linear equation can be derived in the form of \[\frac{dx}{dy}\] as shown below: \[\frac{dx}{dy}+Px=Q\]

where P and Q is in the form of y or any constant (but not in x)

For solving this type of LDE, we adopt the same procedures as in the previous case.

Step 1: First of all, find I.F. (Intergrating factor)

I.F.\[={{e}^{\int Pdy}}\]

Step 2: Further hence, the general solution will be given by

\[x.I.F=\int{Q.(I.F.)dy}\]

• Equation Reducible to Linear Form (Bernoulli Equation): There are the equations of the type

\[\frac{dy}{dx}+Py=Q.{{y}^{n}}\]                            (1)

Where P and Q are the functions of x, which do not appear directly to be of linear form, but can be easily reduced into the linear form by a suitable transformation.

• Method for solving Bernoulli Equations

Step 1: Divide both sides of equation (1) by\[{{y}^{n}}\], we get

\[\frac{1}{{{y}^{n}}}\frac{dy}{dx}+P.\frac{1}{{{y}^{n-1}}}=Q\]             (2)

Step 2: Then put \[\frac{1}{{{y}^{n-1}}}={{y}^{-n+1}}=z\]

\[\Rightarrow \ \ \ \frac{dz}{dx}=(-n+1).{{y}^{-n}}.\frac{dy}{dx}\]                     \[\Rightarrow \]   \[\frac{(1-n)}{{{y}^{n}}}.\frac{dy}{dx}=\frac{dz}{dx}\]          

Then equation (2) becomes

\[\frac{1}{1-n}.\frac{dz}{dx}+Pz=Q\]     \[\Rightarrow \]   \[\frac{dz}{dx}+(1-n).\]\[Pz=(1-n).\]Q

This equation is linear in z. And hence it can be solved by the methods discussed in previous section.

Solved Example

1. Solve the equation: \[\frac{dy}{dx}+\frac{u}{x}=\frac{{{y}^{2}}}{{{x}^{2}}}\]

Sol.      Dividing both sides by \[{{y}^{2}}\], we get

\[\frac{1}{{{y}^{2}}}.\frac{dy}{dx}+\frac{1}{xy}=\frac{1}{{{x}^{2}}}\]                  (1)

Now put \[\frac{1}{y}=z,\] it gives

\[\frac{dz}{dx}=\frac{-1}{{{y}^{2}}}.\frac{dy}{dx}\]

Now the equation (1) reduces to

\[\frac{-dz}{dx}+\frac{z}{x}=\frac{1}{{{x}^{2}}}\]     \[\Rightarrow \frac{dz}{dx}-\frac{z}{x}=\frac{-1}{{{x}^{2}}}\]

which is the linear equation in the form of z.

Here, \[P=\frac{-1}{x}\]and \[Q=\frac{1}{{{x}^{2}}}\]

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

Hence the general solution will be

\[z.\frac{1}{x}=\int{\frac{-1}{{{x}^{2}}}.\frac{1}{x}.dx}\]

\[\Rightarrow \frac{z}{x}=\int{\frac{-1}{{{x}^{3}}}.dx}=\frac{-1}{2}.\frac{1}{{{x}^{2}}}+c\]

\[\Rightarrow \frac{z}{x}=\frac{1}{2{{x}^{2}}}+c\]\[\Rightarrow \frac{1}{yx}=\frac{1}{2{{x}^{2}}}+c\]

\[\Rightarrow 2x=y+2{{x}^{2}}yc\]\[\Rightarrow y=\frac{2x}{1+k{{x}^{2}}}\]

which is the required solution.

2. Solve: the equation: \[x\frac{dy}{dx}+2y={{x}^{2}}\log x\]

Sol.      Given Differential Equation is

\[x\frac{dy}{dx}+2y={{x}^{2}}\log x\]

Dividing by x on both sides, we get

\[\frac{dy}{dx}+\frac{2y}{x}=x\log x\]

Here, \[P=\frac{2}{x}\]and \[Q=x\log x\]

\[\therefore \] Integrating factor, I.F.\[={{e}^{\int P.dx}}\]

\[={{e}^{\int \frac{2}{x}.dx}}\]\[={{e}^{\log {{x}^{2}}}}\]\[={{x}^{2}}\]

Hence the general solution will be

\[y(I.F.)=\int{Q.(I.F.)dx}\]

\[\Rightarrow \]\[y{{x}^{2}}=\int{(x\log x){{x}^{2}}.dx}\]          \[\Rightarrow y{{x}^{2}}=\int{{{x}^{3}}.\log x.dx}\]

\[\Rightarrow y{{x}^{2}}=\log x\int{{{x}^{3}}dx-\int{\left( \frac{d(\log x)}{dx}\int{{{x}^{3}}dx} \right)dx}}\]

(Integrating by parts)

\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\int{\left( \frac{1}{x}.\frac{{{x}^{4}}}{4} \right)dx}\]            \[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\int{\frac{{{x}^{3}}}{4}dx}\]

\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}+c\]             \[\Rightarrow y=\frac{{{x}^{2}}\log x}{4}-\frac{{{x}^{2}}}{16}+c\]

\[\Rightarrow y=\frac{{{x}^{2}}}{16}(4\log x-1)+c\]