Let a curve y = f(x) pass through the points (0, 5) and . If the curve satisfies the differential equation , then k is equal to [2025]

Question

Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]

Smart Achievers · Accepted Answer

(3)

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

$\Rightarrow \frac{d y}{d x} = \frac{2 (3 + y) e^{2 x}}{(7 + e^{2 x})}$

$\Rightarrow \frac{d y}{d x} - \frac{2 e^{2 x}}{(7 + e^{2 x})} \times y = \frac{6 e^{2 x}}{(7 + e^{2 x})}$

$∴ I . F . = e^{- \int (\frac{2 e^{2 x}}{7 + e^{2 x}}) d x} = \frac{1}{7 + e^{2 x}}$

$∴ y \times \frac{1}{7 + e^{2 x}} = \int \frac{6 e^{2 x}}{{(7 + e^{2 x})}^{2}} d x \Rightarrow \frac{y}{7 + e^{2 x}} = \frac{- 3}{(7 + e^{2 x})} + c$

Since, f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ .

Then, $\frac{5}{8} = \frac{- 3}{8} + c \Rightarrow c = 1$

$∴ y = - 3 + (7 + e^{2 x}) \Rightarrow y = e^{2 x} + 4$

$∴ k = e^{2 \log_{e} 2} + 4$

$\Rightarrow k = e^{\log_{e} 2^{2}} + 4 \Rightarrow k = 4 + 4 = 8$

Solution

Q.
Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]

1 16

2 32

3 8

4 4

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Solution

Q. Let a curve y = f(x) pass through the points (0, 5) and (loge2,k). If the curve satisfies the differential equation 2(3+y)e2xdx–(7+e2x)dy=0, then k is equal to [2025]

1 16

2 32

3 8

4 4

Want Us to Email you About Special Offers & Updates?

Q.
Let a curve y = f(x) pass through the points (0, 5) and $(\log_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to [2025]