Let a curve y=f(x) pass through the points (0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the differential equation: $2(3+y) e^{2x} dx = (7+e^{2x}) dy$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{2(3+y) e^{2x}}{7+e^{2x}}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given the differential equation: $2(3+y) e^{2x} dx = (7+e^{2x}) dy$. Rearranging the terms,we get: $\frac{dy}{dx} = \frac{2(3+y) e^{2x}}{7+e^{2x}}$. Separating the variables: $\frac{dy}{3+y} = \frac{2e^{2x}}{7+e^{2x}} dx$. Integrating both sides: $\int \frac{dy}{3+y} = \int \frac{2e^{2x}}{7+e^{2x}} dx$. Let $u = 7+e^{2x}$,then $du = 2e^{2x} dx$. So,$\ln|3+y| = \ln|7+e^{2x}| + C$. This can be written as $3+y = C(7+e^{2x})$. Since the curve passes through $(0,5)$,we substitute $x=0$ and $y=5$: $3+5 = C(7+e^0) \Rightarrow 8 = 8C \Rightarrow C=1$. Thus,the equation of the curve is $3+y = 7+e^{2x}$,which simplifies to $y = e^{2x} + 4$. Now,for the point $(\log_e 2, k)$,we substitute $x = \log_e 2$: $k = e^{2 \log_e 2} + 4 = e^{\log_e 4} + 4 = 4 + 4 = 8$.

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

Similar Questions

The equation of the curve passing through $(1, 0)$ and having a slope of $\frac{y - 1}{x^2 + x}$ is:

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ is

The solution of the differential equation $\frac{dy}{dx} = \sin(x + y) + \cos(x + y)$ is

Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0$.

Find the equation of a curve passing through the point $(-2, 3),$ given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module