Question 2

y = x2 + 2x + C y' - 2x - 2 = 0   

Answer

y = x2 + 2x + C

Differentiating both sides of this equation with respect to x, we get:

\begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}

=> y' = 2x + 2

Substituting the value of y' in the given differential equation, we get:

L.H.S. = y' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

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