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Question
Show that y = C x + 2C2 is a solution of the differential equation \[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0.\]
Solution
We have,
\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0 . . . . . \left( 1 \right)\]
Now,
y = C x + 2C2
\[\Rightarrow\frac{dy}{dx}=C\]
\[\text{Putting }\frac{dy}{dx} = C\text{ and }y = Cx + 2 C^2\text{ in (1), we get}\]
\[\text{LHS }= 2 \left( C \right)^2 + x\left( C \right) - \left( Cx + 2 C^2 \right)\]
\[ = 2 C^2 + xC - xC - 2 C^2 \]
\[ = 0 =\text{ RHS}\]
Thus, y = C x + 2C2 is the solution of the given differential equation.
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