UPSC 2023 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Linear first-order · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Obtain the solution of the initial-value problem $\dfrac{dy}{dx}-2xy=2$ , $y(0)=1$ in the form

y=e^{x^2}\bigl[1+\sqrt\pi\,\operatorname{erf}(x)\bigr].

Linear ODE with integrating factor $e^{-x^2}$ ; integrate the RHS to introduce the error function $\operatorname{erf}$ .

Step 1 — Linear ODE; find integrating factor.

The ODE is $y'-2xy=2$ , a first-order linear ODE.

Integrating factor: $\mu(x)=\exp(-\int 2x\,dx)=e^{-x^2}$ .

Multiply through:

e^{-x^2}\frac{dy}{dx}-2x e^{-x^2}y=2 e^{-x^2}\Rightarrow \frac{d}{dx}\bigl(e^{-x^2}y\bigr)=2e^{-x^2}.

Step 2 — Integrate.

e^{-x^2}y=\int 2 e^{-x^2}\,dx+C.

Recall $\operatorname{erf}(x)=\dfrac{2}{\sqrt\pi}\int_0^x e^{-t^2}\,dt$ . So $\int 2e^{-x^2}\,dx=\sqrt\pi\,\operatorname{erf}(x)+\text{const}$ .

Therefore $e^{-x^2}y=\sqrt\pi\,\operatorname{erf}(x)+C$ .

Step 3 — Apply initial condition $y(0)=1$ .

At $x=0$ : $e^0\cdot 1=1=\sqrt\pi\cdot\operatorname{erf}(0)+C=0+C$ , so $C=1$ .

Hence

y=e^{x^2}\bigl[\sqrt\pi\,\operatorname{erf}(x)+1\bigr]=\boxed{\;e^{x^2}\bigl[1+\sqrt\pi\,\operatorname{erf}(x)\bigr].\;}