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Variable Separable

Some differential equations can be solved by the method of separation of variables.

(dy/dx) = f(x, y) is a first order and first degree differential equation. If f(x, y) can be written as h(x).g(y), where h(x) is a function of x and g(y) is a function of y, then we can separate h(x) and g(y).

Solutions of differential equations of the type (dy/dx) = f(ax + by + c):

Step1: Substitute ax + by + c = v in the given differential equation and change the differential coefficients accordingly.

Step2: Obtain the relation f(v)dv = g(x)dx

Step3: Integrate the obtained equation to get a relation between x, v and arbitrary constants.

Step4: Substitute the value of v in the relation obtained in step-3 to get a relation between x, y and arbitrary constants.

Keywords: Variable Separable method, Solutions of differential equations

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Solved Board Paper

Q1
Marks:1
Answer:

Explanation:

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Q2
Marks:1
Answer:

Explanation:

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Q3
The general solution of the differential equation

(1 + x²) dy = (1 + y²) dx is
Marks:1
Answer:

tan^-1y = tan^-1x + C.

Explanation:

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Q4
The general solution of the differential equation y(x²y + e^x)dx – e^xdy = 0 is
Marks:1
Answer:

x³y + 3e^x = Cy

Explanation:

$We have, y (x^{2} y + e^{x}) d x - e^{x} d y = 0$

$\Rightarrow e^{x} \frac{d y}{d x} = x^{2} y^{2} + y e^{x}$

$Dividing by y^{2} e^{x}, we get$

$\frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} = x^{2} e^{- x}$

$Putting \frac{1}{y} = v, we get$

$- \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d v}{d x}$

$Thus, we have \frac{d y}{d x} + v = - x^{2} e^{- x} which is a linear differential equation .$

$I . F . = e^{\int 1 d x} = e^{x}$

Hence, the solution is

$v \cdot e^{x} = - \int x^{2} e^{- x} \cdot e^{x} d x + \frac{C}{3}$

$or \frac{1}{y} e^{x} = - \frac{x^{3}}{3} + \frac{C}{3} \Rightarrow x^{3} y + 3 e^{x} = C y$

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Q5
The differential equation, (e^x + 1)ydy = (y + 1)e^xdx has the solution
Marks:1
Answer:

(1 + y)(e^x + 1) = Ce^y.

Explanation:

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