The general solution of the differential equa | vedclass

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

Question

(C) Given differential equation: $\frac{dy}{dx} = y 	an x - y^2 \sec x$ Divide by $y^2$: $\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{y} 	an x = -\sec x$

Vedclass · Accepted Answer

$\sec x = (c + 	an x) y$,where $c$ is constant of integration.

Vedclass · Answer

(C) Given differential equation: $\frac{dy}{dx} = y 	an x - y^2 \sec x$ Divide by $y^2$: $\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{y} 	an x = -\sec x$ Let $v = -\frac{1}{y}$,then $\frac{dv}{dx} = \frac{1}{y^2} \frac{dy}{dx}$. Substituting this into the equation: $\frac{dv}{dx} + v 	an x = -\sec x$. This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = 	an x$ and $Q(x) = -\sec x$. Integrating factor ($I$.$F$.) $= e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$. The solution is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$. $v \sec x = \int -\sec x \cdot \sec x dx + c = -\int \sec^2 x dx + c$. $v \sec x = -	an x + c$. Substituting $v = -\frac{1}{y}$: $-\frac{1}{y} \sec x = -	an x + c$. Multiplying by $-y$: $\sec x = y(	an x - c)$. Replacing $-c$ with a new constant $c$: $\sec x = y(	an x + c)$.

The general solution of the differential equation $\frac{dy}{dx} = y \tan x - y^2 \sec x$ is

Similar Questions

The solution of the differential equation $y^{\prime} = \frac{1}{e^y - x}$ is

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y = \frac{x+3}{x+1}$,where $x > -1$,which passes through the point $(0,1)$. Then $y(1)$ is equal to:

If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x=2 x \sec x$ and $y(0)=0$,then which of the following is true?

The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

Which of the following equation$(s)$ is/are linear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module