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

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#### Solution

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,

\[y = 1\text{ at }x = 2\]

\[ \therefore 2 = 2 + C\]

\[ \Rightarrow C = 0\]

\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]

\[x = 2 y^2 \]

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

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