Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
Let $x = x(y)$ be the solution of the differential equation $y^2\,dx+\left(x-\dfrac{1}{y}\right)dy = 0$. If $x(1) = 1$, then $x\!\left(\dfrac{1}{2}\right)$ is:
$\dfrac{1}{2}+e$
$3+e$
$3-e$
$\dfrac{3}{2}+e$
Step-by-Step Solution
Key Concept: Rearrange as $\dfrac{dx}{dy}+\dfrac{x}{y^2} = \dfrac{1}{y^3}$; integrating factor $= e^{\int 1/y^2\,dy} = e^{-1/y}$; integrate by parts using the substitution $t=-1/y$.
Rearranging: $\dfrac{dx}{dy}+\dfrac{x}{y^2} = \dfrac{1}{y^3}$. I.F. $= e^{-1/y}$.
$xe^{-1/y} = \int\dfrac{e^{-1/y}}{y^3}dy$. Let $t=-1/y$: $= -\int e^t t\,dt = -e^t(t-1)+C = e^{-1/y}\!\left(\dfrac{1}{y}+1\right)\cdot(-1)+C$.
Actually: $xe^{-1/y} = -e^{-1/y}\!\left(\dfrac{-1}{y}-1\right)+C \Rightarrow x = \dfrac{1}{y}+1-e^{1+1/y}\cdot e^{-1/y}\cdot e$...
Cleaning up: $x = \dfrac{1}{y}+1-e^{1-1+1/y}$. Using $x(1)=1$: $C=-e^{-1}$. Hence $x = \dfrac{1}{y}+1-e^{-1+1/y}$.
$$x\!\left(\frac{1}{2}\right) = 2+1-e^{-1+2} = 3-e.$$
Correct Answer: 3