Differential Equations
ODE — Bernoulli — Finding α+β-γ
nta_pyq_2023_jan
Grade None
Question:
Let $\alpha x=\exp(x^\beta y^\gamma)$ be the solution of the differential equation $2x^2y\,dy-(1-xy^2)\,dx=0$, $x>0$, $y(2)=\sqrt{\log_e 2}$. Then $\alpha+\beta-\gamma$ equals:
Step-by-Step Solution
Key Concept: Let $t=xy^2$: $dt=y^2dx+2xy\,dy=(y^2+2xy\cdot\frac{dy}{dx})dx$. From ODE: $2x^2y\,dy=(1-xy^2)dx\Rightarrow 2xy\,dy/dx=(1-t)/x$. So $dt/dx=y^2+(1-t)/x=t/x+(1-t)/x=1/x$. $t=\ln x+C$.
$\alpha+\beta-\gamma=1$.
Correct Answer: 1