Question

Solve the differential equation \[\left( x + 2 y^2 \right)\frac{dy}{dx} = y\], given that when x = 2, y = 1.

Answer 1

We have, 
\[\left( x + 2 y^2 \right)\frac{dy}{dx} = y\]
\[ \Rightarrow \frac{dx}{dy} = \frac{1}{y}\left( x + 2 y^2 \right) \]
\[ \Rightarrow \frac{dx}{dy} - \frac{1}{y}x = 2y . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dx}{dy} + Px = Q\]
where
\[P = - \frac{1}{y}\]
\[Q = 2y\]
\[ \therefore I . F . = e^{ \int P dy } \]
\[ = e^{- \int\frac{1}{y}dy} \]
\[ = e^{- \log y} = \frac{1}{y}\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }\frac{1}{y},\text{ we get }\]
\[\frac{1}{y}\left( \frac{dx}{dy} - \frac{1}{y}x \right) = \frac{1}{y} \times 2y\]
\[ \Rightarrow \frac{1}{y}\frac{dx}{dy} - \frac{1}{y^2}x = 2\]
Integrating both sides with respect to y, we get
\[x\frac{1}{y} = \int 2dy + C\]
\[ \Rightarrow x\frac{1}{y} = 2y + C\]
\[ \Rightarrow x = 2 y^2 + Cy . . . . . \left( 2 \right)\]
Now, 
\[y = 1\text{ at }x = 2\]
\[ \therefore 2 = 2 + C\]
\[ \Rightarrow C = 0\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[x = 2 y^2 \]
\[\text{ Hence, }x = 2 y^2\text{ is the required solution .}\]