Question

If y = e^x (sin x + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?

Answer 1

Here,

\[y = e^x \left( \sin x + \cos x \right)\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = e^x \left( \sin x + \cos x \right) + e^x \left( \cos x - \sin x \right) = 2 e^x \cos x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = 2 e^x \cos x - 2 e^x \sin x\]

\[\text { Now, }\]

\[\text { LHS  }= \frac{d^2 y}{d x^2} - 2\frac{d y}{d x} + 2y\]

\[ = 2 e^x \cos x - 2 e^x \sin x - 4 e^x \cos x + 2 e^x \left( \sin x + \cos x \right)\]

\[ = 0 = \text { RHS }\]

Hence proved.