Solve the following differential equation (x+2y^2) dy dx =y, give… - Vidyadip

Question

Solve the following differential equation $(\text{x}+2\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y},$ given that when x = 2, y = 1.

✓

Answer

We have,

$(\text{x}+2\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}$

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

$\Rightarrow\ \frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}}\text{x}=2\text{y}\ \dots(1)$

Clearly, it is a linear differential equation of the form

$\frac{\text{dx}}{\text{dy}}+\text{Px = Q}$

where

$\text{P}=-\frac{1}{\text{y}}$

$\text{Q}=2\text{y}$

$\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$

$=\text{e}^{-\int\frac{1}{\text{y}}\text{dy}}$

$=\text{e}^{-\log\text{y}}$

$=\frac{1}{\text{y}}$

Multiplying both sides of (1) by $\frac{1}{\text{y}},$ we get

$\frac{1}{\text{y}}\Big(\frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}}\text{x}\Big)=\frac{1}{\text{y}}\times2\text{y}$

$\Rightarrow\ \frac{1}{\text{y}}\frac{\text{dx}}{\text{dy}}-\frac{1}{\text{y}^2}\text{x}=2$

Integrating both sides with respect to y, we get

$\text{x}\frac{1}{\text{y}}=\int2\text{dy + C}$

$\Rightarrow\ \text{x}\frac{1}{\text{y}}=2\text{y + C}$

$\Rightarrow\ \text{x}=2\text{y}^2+\text{Cy}\ \dots(2)$

Now,

$\text{y}=1$ at $\text{x}=2$

$\therefore\ 2=2+\text{C}$

$\Rightarrow\ \text{C}=0$

Putting the value of C in (2), we get

$\text{x}=2\text{y}^2$

Hence, $\text{x}=2\text{y}^2$ is the required solution.

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