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Differential Equations question

2025 · 23 Jan · Shift 1 · Q26

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Mathematics
Differential Equations
2025 · 23 Jan · Shift 1 · Q26

Differential Equations question

2025 · 23 Jan · Shift 1 · Q26

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let a curve

y=f(x)

pass through the points

(0,5)

and

\left(\log _e 2, k\right)

. If the curve satisfies the differential equation

2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0

, then

k

is equal to

A
32
B
8
C
4
D
16

View written solutionFree

Correct answer: B

Write the differential equation in derivative form

Given $2(3+y)e^{2x}\,dx-(7+e^{2x})\,dy=0$

Rearrange: $2(3+y)e^{2x}\,dx=(7+e^{2x})\,dy$

So, $\frac{dy}{dx}=\frac{2(3+y)e^{2x}}{7+e^{2x}}$

This is separable.

Separate variables

$\frac{dy}{3+y}=\frac{2e^{2x}}{7+e^{2x}}\,dx$

Integrate both sides

Left side: $\int \frac{dy}{3+y}=\ln|3+y|$

Right side: $\int \frac{2e^{2x}}{7+e^{2x}}\,dx$

Let $t=7+e^{2x} \implies dt=2e^{2x}\,dx$

Hence, $\int \frac{2e^{2x}}{7+e^{2x}}\,dx=\int \frac{dt}{t}=\ln|t|=\ln(7+e^{2x})$

Therefore, $\ln|3+y|=\ln(7+e^{2x})+C$

So, $3+y=C(7+e^{2x})$

Use the point $(0,5)$ to find the constant

At $x=0$ , $y=5$ : $3+5=C(7+e^0)$ $8=C(7+1)=8C$ $C=1$

Thus the curve is $3+y=7+e^{2x}$ $y=4+e^{2x}$

Use the point $\left(\ln 2,k\right)$

At $x=\ln 2$ : $k=4+e^{2\ln 2}$

Now, $e^{2\ln 2}=e^{\ln 4}=4$

So, $k=4+4=8$

Compare with options

$k=8$

So the correct option is B.

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