QJEEM23Mac404: JEE Main 2023 Mathematics - Class 12 JEEM23Mac404 Solution: Let be the solution of the differential equation with . Then is …

Question

Verified Solution • 2025–26 Session moderate

Let

x=x(y)

be the solution of the differential equation

2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y&gt;-1

with

x\left(e^{4}-2\right)=1

. Then

x\left(e^{9}-2\right)

is equal to :

Accepted Answer

\begin{aligned}&amp; 2(y+2) \ln (y+2) d x+(x+4-2 \ln (y+2)) d y=0 \\\\&amp; 2 \ln (y+2)+(x+4-2 \ln (y+2)) \frac{1}{y+2} \cdot \frac{d y}{d x}=0 \\\\&amp; \text { let, } \ln (y+2)=t \\\\&amp; \frac{1}{y+2} \cdot \frac{d y}{d x}=\frac{d t}{d x} \\\\&amp; 2 t+(x+4-2 t) \cdot \frac{d t}{d x}=0 \\\\&amp; (x+4-2 t) \frac{d t}{d x}=-2 t \\\\&amp; \frac{d x}{d t}=\frac{2 t-4-x}{2 t} \\\\&amp; \frac{d x}{d t}+\frac{x}{2 t}=\frac{2 t-4}{2 t}\end{aligned}

\begin{aligned}&amp; x \cdot t^{1 / 2}=\int \frac{2 t-4}{2 t} \cdot t^{1 / 2} \cdot d t \\\\&amp; x \cdot t^{1 / 2}=\int\left(t^{1 / 2}-\frac{2}{t^{1 / 2}}\right) \cdot d t \\\\&amp; =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\\\&amp; x \cdot t^{\frac{1}{2}}=\frac{2 t^{\frac{3}{2}}}{3}-4 \mathrm{t}^{\frac{1}{2}}+C \\\\&amp; x=\frac{2}{3} \cdot \mathrm{t}-4+\mathrm{C} \cdot \mathrm{t}^{\frac{-1}{2}} \\\\&amp; x=\frac{2}{3} \ln (\mathrm{y}+2)-4+C \cdot(\ln (\mathrm{y}+2))^{\frac{-1}{2}}\end{aligned}

Put

y=e^4-2, x=1

\begin{aligned}&amp; 1=\frac{2}{3} \times 4-4+C \times \frac{1}{2} \\\\&amp; \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \\\\&amp; C=\frac{14}{3} \\\\&amp; x=\frac{2}{3} \times 9-4+\frac{14}{3} \times \frac{1}{3} \\\\&amp; =2+\frac{14}{9} \\\\&amp; =\frac{32}{9}\end{aligned}

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