Question

Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xe^x + e^x

Answer 1

We have,

y = xe^x + e^x                .....(1)

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = x e^x + e^x + e^x \]

\[ \Rightarrow \frac{dy}{dx} = x e^x + 2 e^x ...........(2)\]

Differentiating both sides of (2) with respect to x, we get

\[\frac{d^2 y}{d x^2} = x e^x + e^x + 2 e^x \]

\[ \Rightarrow \frac{d^2 y}{d x^2} = x e^x + 3 e^x \]

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

\[ \Rightarrow \frac{d^2 y}{d x^2} = 2\frac{dy}{dx} - y ...........\left[\text{Using (1) and (2)}\right]\]

\[ \Rightarrow \frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0\]

\[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0\]

It is the given differential equation.

Thus, y = xe^x + e^x satisfies the given differential equation.

Also, when \[x = 0, y = 0 + 1 = 1,\text{ i.e. }y(0) = 1\]

And, when \[x = 0, y' = 0 + 2 = 2,\text{ i.e. }y'(0) = 2\]

Hence, y = xe^x + e^x is the solution to the given initial value problem.