Differential Equations — Variable Separable
PYP_JEE_ADV_2026_P2
Grade None
Question:
Let $y:(-\infty,\infty)\to(0,\infty)$ be the solution of the differential equation
$$\frac{dy}{dx}=\frac{e^{5x}y^3+y^3}{e^x+e^xy^4},$$
satisfying $y(0)=\dfrac{1}{\sqrt{2}}$. Then the value of $y(\log_e 2)$ is
$\sqrt{\dfrac{5+\sqrt{35}}{2}}$
$\sqrt{\dfrac{7+\sqrt{53}}{2}}$
$\dfrac{7+\sqrt{53}}{2}$
$\dfrac{5+\sqrt{35}}{2}$
Step-by-Step Solution
Key Concept: After separation and integration, the implicit equation $\frac{y^2-y^{-2}}{2}=F(x)$ leads to a quadratic in $y^2$. Only the positive root is valid since $y>0$.
**Step 1: Separate variables**
$\frac{dy}{dx}=\frac{y^3(e^{5x}+1)}{e^x(1+y^4)}$. Rearranging: $\frac{1+y^4}{y^3}dy=(e^{4x}+e^{-x})dx$.
**Step 2: Integrate both sides**
$\int(y^{-3}+y)dy=\int(e^{4x}+e^{-x})dx$. $-\dfrac{1}{2y^2}+\dfrac{y^2}{2}=\dfrac{e^{4x}}{4}-e^{-x}+C$.
**Step 3: Apply initial condition $y(0)=1/\sqrt{2}$**
$y^2=1/2$: $-1+1/2 = 1/4-1+C \Rightarrow -1/2=-3/4+C \Rightarrow C=1/4$. Wait: $-\frac{1}{2\cdot(1/2)}+\frac{1/2}{2}=\frac{1}{4}-1+C \Rightarrow -1+\frac{1}{4}=-\frac{3}{4}=\frac{1}{4}-1+C \Rightarrow C=0$.
**Step 4: Evaluate at $x=\ln 2$**
$\frac{y^2-y^{-2}}{2}=\frac{e^{4\ln2}}{4}-e^{-\ln2}=4-\frac{1}{2}=\frac{7}{2}$. So $y^4-1=7y^2 \Rightarrow y^4-7y^2-1=0$. $y^2=\frac{7+\sqrt{53}}{2}$ (positive root). $y=\sqrt{\dfrac{7+\sqrt{53}}{2}}$.
Correct Answer: B