Differential Equations
First Order ODE
MMTS_Full_Test_06
Grade 12
Question:
Let $y=f(x)$ satisfy $\dfrac{dy}{dx}=2xe^{-y}$, $\forall x\in\mathbb{R}$. If $y'(1)=1$, then the number of solutions of $f(x)=f'(x)$ in $(0,\infty)$ is
Step-by-Step Solution
Key Concept: Solve ODE: $e^y dy=2x\,dx\Rightarrow e^y=x^2+C$
$e^y=x^2+C$. $y'=2xe^{-y}=\frac{2x}{x^2+C}$. At $x=1$: $y'=1\Rightarrow\frac{2}{1+C}=1\Rightarrow C=1$. $y=\ln(x^2+1)$, $y'=\frac{2x}{x^2+1}$. Solve $\ln(x^2+1)=\frac{2x}{x^2+1}$: 1 solution in $(0,\infty)$.
Correct Answer: 1