Question

Let \( \varphi : \mathbb{R} \to \mathbb{R} \) be the solution of the differential equation \[ \frac{dy}{dx} + 2xy = 2 + 4x^2, \] satisfying \( \varphi(0) = 0 \). Then, the value of \( \varphi(2) \) is equal to ............. (rounded off to two decimal places).

Hint:

For first-order linear differential equations, find the integrating factor and multiply both sides of the equation to make the left-hand side a derivative.

Answer 1

Step 1: Solve the differential equation.
This is a first-order linear differential equation. The standard form is: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = 2x \) and \( Q(x) = 2 + 4x^2 \). The integrating factor is: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 2x \, dx} = e^{x^2}. \] Step 2: Multiply the equation by the integrating factor.
Multiply both sides of the differential equation by \( e^{x^2} \): \[ e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y = (2 + 4x^2) e^{x^2}. \] The left-hand side is the derivative of \( e^{x^2} y \), so the equation becomes: \[ \frac{d}{dx} \left( e^{x^2} y \right) = (2 + 4x^2) e^{x^2}. \] Step 3: Integrate both sides.
Integrating both sides: \[ e^{x^2} y = \int (2 + 4x^2) e^{x^2} \, dx. \] We use the fact that: \[ \int (2 + 4x^2) e^{x^2} \, dx = e^{x^2} + C. \] Thus: \[ e^{x^2} y = e^{x^2} + C \implies y = 1 + C e^{-x^2}. \] Step 4: Apply the initial condition.
Use \( \varphi(0) = 0 \) to find \( C \): \[ 0 = 1 + C e^0 \implies C = -1. \] Thus, the solution is: \[ y = 1 - e^{-x^2}. \] Step 5: Evaluate \( \varphi(2) \).
Finally, compute \( \varphi(2) \): \[ \varphi(2) = 1 - e^{-4} \approx 1 - 0.0183 = 0.9817. \] Final Answer: \[ \boxed{0.98}. \]