Differential Equations: Numerical Problems Set (11)

Question

Solve the linear differential equation $\dfrac{dy}{dx} + 2y = e^{-x}$, with $y(0) = 1$.

Solution — Step by Step

Why This Works

The integrating factor turns the LHS into a perfect derivative of a product. After multiplying by $e^{\int P \, dx}$, the equation becomes $(y \cdot \text{IF})' = Q \cdot \text{IF}$, which is directly integrable.

This trick works for any linear first-order ODE — find $P(x)$, compute the IF, multiply, integrate. No guessing.

Alternative Method

Try $y = u(x) e^{-2x}$ as an ansatz (this is the form of the general solution of the homogeneous part).

Compute $y' = u' e^{-2x} - 2u e^{-2x}$. Substitute:

$u' e^{-2x} - 2u e^{-2x} + 2u e^{-2x} = e^{-x}$

$u' e^{-2x} = e^{-x} \implies u' = e^x \implies u = e^x + C$

So $y = (e^x + C)e^{-2x} = e^{-x} + Ce^{-2x}$. Same.

Common Mistake

Students forget to multiply both sides by the integrating factor before recognising the LHS as a perfect derivative. They write $\dfrac{d}{dx}(y \cdot e^{2x})$ on the LHS but leave $Q(x) = e^{-x}$ alone on the RHS. Wrong — the IF must multiply both sides.

Another classic: integrating $e^{2x}$ as $e^{2x}/2$ versus $\int 2 \, dx = 2x$ versus $\int 2 \, dx$ becoming the IF exponent. Keep them straight: the IF is $e^{\int P \, dx}$, not $\int P \, dx$ itself.