Question

Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = e^x + e^2x

Answer 1

We have,

\[y = e^x + e^{2x}.............(1)\]

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

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

Differentiating both sides of (2) with respect to x, we get 
\[\frac{d^2 y}{d x^2} = e^x + 4 e^{2x} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 3\left( e^x + 2 e^{2x} \right) - 2\left( e^x + e^{2x} \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 3\frac{dy}{dx} - 2y ..........\left[\text{Using (1) and (2)}\right]\]

\[ \Rightarrow \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0\]

\[\Rightarrow \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0\]

It is the given differential equation.

Therefore, y = ex + e2x satisfies the given differential equation.

Also, when \[x = 0; y = e^0 + e^0 = 1 + 1,\text{ i.e. }y(0) = 2\]

And, when \[x = 0; y' = e^0 + 2 e^0 = 1 + 2,\text{ i.e. }y'(0) = 3\]
Hence, `y = e^x + e^(2x)` is the solution to the given initial value problem.