AIEEE 2011 (Offline)Question 1
If $\frac{dy}{dx}=y+3>0$ and $y(0)=2$, then $y(\ln2)$ is equal to :
Differential Equation
Solution
Concept. $\frac{d(y+3)}{dx}=y+3$, so $y+3$ grows like $e^x$. Formulas used. (1) $\dfrac{dz}{dx}=z\Rightarrow z=z_0e^x$Step 1 - Solve. By (1), $y+3=5e^x$ (since $y(0)+3=5$). Step 2 - At $x=\ln2$. $y=10-3=7$. Answer: (D)