Solve the following differential equation (x+2y^2) dy dx =y, given that when x = 2, y = 1.

We have, (x+2y^2) dy dx =y dx dy =1 y (x+2y^2) dx dy -1 y x=2y (1) Clearly, it is a linear differential equation of the form dx dy +Px = Q where P=-1 y Q=2y I.F. =e^ =e^ - 1 y dy =e^ - =1 y Multiplying both sides of (1) by 1 y , we get 1 y ( dx dy -1 y x )=1 y 2y 1 y dx dy -1 y^2 x=2 Integrating both sides with respect to y, we get x1 y = 2dy + C x1 y =2y + C x=2y^2+Cy (2) Now, y=1 at x=2 2=2+C C=0 Putting the value of C in (2), we get x=2y^2 Hence, x=2y^2 is the required solution.

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

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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.

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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.