The solution curve of the differential equati | vedclass

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

Question

(A) Given differential equation: $(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$.Separate the variables:$\frac{1+y^{2}}{y^{2}} dy = \frac{1}{1+e^{-x}} dx$

Vedclass · Accepted Answer

$y^{2}=1+y \log _{e}\left(\frac{1+e^{x}}{2}\right)$

Vedclass · Answer

(A) Given differential equation: $(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$.Separate the variables:$\frac{1+y^{2}}{y^{2}} dy = \frac{1}{1+e^{-x}} dx$$\Rightarrow (y^{-2}+1) dy = \frac{e^{x}}{e^{x}+1} dx$.Integrating both sides:$\int (y^{-2}+1) dy = \int \frac{e^{x}}{e^{x}+1} dx$$-y^{-1} + y = \ln(e^{x}+1) + C$$y - \frac{1}{y} = \ln(e^{x}+1) + C$.Since the curve passes through $(0,1)$,substitute $x=0$ and $y=1$:$1 - \frac{1}{1} = \ln(e^{0}+1) + C$$0 = \ln(2) + C \Rightarrow C = -\ln(2)$.Substituting $C$ back into the equation:$y - \frac{1}{y} = \ln(e^{x}+1) - \ln(2)$$y - \frac{1}{y} = \ln\left(\frac{e^{x}+1}{2}\right)$Multiply by $y$:$y^{2} - 1 = y \ln\left(\frac{1+e^{x}}{2}\right)$$y^{2} = 1 + y \ln\left(\frac{1+e^{x}}{2}\right)$.

The solution curve of the differential equation,$(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$,which passes through the point $(0,1)$,is

Similar Questions

The solution of the differential equation $(1 - x^2)(1 - y)dx = xy(1 + y)dy$ is

If $f(x)$ is a function such that $f^{\prime}(x)=\sqrt{f^2(x)-1}$ and $f(0)=1$,then $f(1)=$

If $y = f(x)$ is a monotonically increasing function such that $(\frac{dy}{dx})^2 = 6 - \frac{dy}{dx}$ and $y(0) = 5$, then $y(3) = ...$

Find a particular solution satisfying the given condition:
$x(x^{2}-1) \frac{dy}{dx}=1; y=0$ when $x=2$

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module