Name: Exact Differential Equations
Uploaded: 2019-07-19
Duration: 14 min 3 s
Description: An exact differential equation is a certain kind of ordinary differential equation that is widely used in mathematics. Watch this video and learn about the e...

Question

Use the substitution

y^{2} = a - x

to reduce the equation

y^{3} \cdot \frac{d y}{d x} + x + y^{2} = 0

to homogeneous form and hence solve it.

(

where

a

is variable

)

Accepted Answer

Given $y^{3} \cdot \frac{d y}{d x} + x + y^{2} = 0$ .

Put $y^{2} = a - x$

$\Rightarrow 2 y \cdot \frac{d y}{d x} = \frac{d a}{d x} - 1$

So $\frac{1}{2} (\frac{d a}{d x} - 1) (a - x) + x + a - x = 0$

$\Rightarrow \frac{1}{2} (\frac{d a}{d x} - 1) = \frac{- a}{a - x}$

$\Rightarrow \frac{d a}{d x} - 1 = \frac{2 a}{x - a}$

$\Rightarrow \frac{d a}{d x} = \frac{2 a}{x - a} + 1$

$∴ \frac{d a}{d x} = \frac{x + a}{x - a} .$

Put $a = v x$

$\Rightarrow \frac{d a}{d x} = v + x \frac{d v}{d x}$

So $v + x \frac{d v}{d x} = \frac{x + v x}{x - v x}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{1 + v}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v$

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

$\Rightarrow \int \frac{(1 - v) d v}{1 + v^{2}} = \int \frac{d x}{x}$

$\Rightarrow \int \frac{d v}{1 + v^{2}} - \int \frac{v d v}{1 + v^{2}} = \int \frac{d x}{x}$

Put $1 + v^{2} = t, \Rightarrow 2 v d v = d t$ in the second integral, to get

$\Rightarrow \int \frac{d v}{1 + v^{2}} - \frac{1}{2} \int \frac{d t}{t} = \int \frac{d x}{x}$

Using $\int \frac{1}{1 + x^{2}} d x = \tan^{- 1} x$ and $\int \frac{1}{x} d x = \ln |x|,$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \ln t = \ln x + c$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \ln |1 + v^{2}| = \ln x + c$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} \ln |1 + \frac{a^{2}}{x^{2}}| = \ln x + c$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} \ln |\frac{x^{2} + a^{2}}{x^{2}}| = \ln x + c$

Using $\ln (\frac{m}{n}) = \ln (m) - \ln (n),$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} [\ln |x^{2} + a^{2}| - \ln x^{2}] = \ln x + c$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} [\ln |x^{2} + a^{2}| - 2 \ln x] = \ln x + c$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} \ln |x^{2} + a^{2}| + \ln x = \ln x + c$

$\Rightarrow \tan^{- 1} \frac{a}{x} - \frac{1}{2} \ln |x^{2} + a^{2}| = c,$ where $a = x + y^{2} .$

Use the substitution $y^{2} = a - x$ to reduce the equation $y^{3} \cdot \frac{d y}{d x} + x + y^{2} = 0$ to homogeneous form and hence solve it. $($ where $a$ is variable $)$

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Use the substitution y2=a-x to reduce the equation y3·dydx+x+y2=0 to homogeneous form and hence solve it. (where a is variable)

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Use the substitution $y^{2} = a - x$ to reduce the equation $y^{3} \cdot \frac{d y}{d x} + x + y^{2} = 0$ to homogeneous form and hence solve it. $($ where $a$ is variable $)$