Differential Equations
Linear ODE — Integrating Factor
nta_pyq_2023_jan
Grade 12
Question:
Let $y=y(x)$ be the solution of the differential equation $x^3\,dy+(xy-1)\,dx=0$, $x>0$, $y\!\left(\dfrac{1}{2}\right)=3-e$. Then $y(1)$ is equal to:
Step-by-Step Solution
Key Concept: Rewrite: $\dfrac{dy}{dx}+\dfrac{y}{x^2}=\dfrac{1}{x^3}$. IF $=e^{\int 1/x^2\,dx}=e^{-1/x}$. $ye^{-1/x}=\int e^{-1/x}/x^3\,dx$. Substitute $t=-1/x$: $=-\int e^t t\,dt=-(e^t t-e^t)+C=e^{-1/x}(1/x+1)+C$.
$y(1)=1$.
Correct Answer: 1