First slide

Methods of solving first order first degree differential equations

Question

The solution of the differential equation $x \frac{d y}{d x} + y = x^{3} y^{6}$ , is

Moderate

A

$x^{7} = 5 y^{5} + C x^{2} y^{5}$
B

$2 x^{7} = 5 y^{5} + C x^{2} y^{5}$
C

$5 x^{7} = 2 y^{5} + C x^{2} y^{5}$
D

$2 x^{7} = 5 y^{5} + C x^{5} y^{2}$

Solution

The given differential equation can be written as
$\frac{1}{y^{6}} \frac{d y}{d x} + \frac{1}{x y^{5}} = x^{2}$
Let $y^{- 5} = 0$ , Then
$- 5 y^{- 6} \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow y^{- 6} \frac{d y}{d x} = - \frac{1}{5} \frac{d v}{d x}$
Substituting these values in the given differential equation, we get
$- \frac{1}{5} \frac{d v}{d x} + \frac{1}{x} v = x^{2}$
$\Rightarrow \frac{d v}{d x} - \frac{5}{x} v = - 5 x^{2}$ …(i)
This is the standard form of the linear differential equation having integrating factor
$I. F. = e^{\int \frac{5}{x} d x} = e^{- 5 \log x} = \frac{1}{x^{5}}$
Multiplying both sides of (i) by I.F. and integrating w.r.t. x, we get
$\begin{array}{l} v \frac{1}{x^{5}} = \int - 5 x^{2} \cdot \frac{1}{x^{5}} d x \\ \Rightarrow \frac{v}{x^{5}} = \frac{5}{2} x^{- 2} + C \end{array}$
$\Rightarrow y^{- 5} x^{5} = \frac{5}{2} x^{- 2} + C$ , which is the required solution.

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