General solution of (x+y)^{2} frac{d y}{d x}= | vedclass

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

Question

(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a 
eq 0$Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$, which implies $\frac{d y}{d x}=\

Vedclass · Accepted Answer

$	an \frac{y+C}{a}=\frac{x+y}{a}$

Vedclass · Answer

(D) Given equation: $(x+y)^{2} \frac{d y}{d x}=a^{2}, a 
eq 0$Let $x+y=t$. Then $1+\frac{d y}{d x}=\frac{d t}{d x}$, which implies $\frac{d y}{d x}=\frac{d t}{d x}-1$.Substituting this into the given equation:$t^{2}(\frac{d t}{d x}-1)=a^{2}$$t^{2} \frac{d t}{d x} = a^{2}+t^{2}$Separating the variables:$\frac{t^{2}}{a^{2}+t^{2}} d t = d x$Integrating both sides:$\int \frac{t^{2}+a^{2}-a^{2}}{t^{2}+a^{2}} d t = \int d x$$\int (1 - \frac{a^{2}}{t^{2}+a^{2}}) d t = x + C'$$t - a^{2} \cdot \frac{1}{a} 	an^{-1}(\frac{t}{a}) = x + C'$$t - a 	an^{-1}(\frac{t}{a}) = x + C'$Substituting $t = x+y$ back:$(x+y) - a 	an^{-1}(\frac{x+y}{a}) = x + C'$$y - C' = a 	an^{-1}(\frac{x+y}{a})$$\frac{y-C'}{a} = 	an^{-1}(\frac{x+y}{a})$$	an(\frac{y-C'}{a}) = \frac{x+y}{a}$Let $C = -C'$. Then the general solution is $	an(\frac{y+C}{a}) = \frac{x+y}{a}$.

General solution of $(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$ is ($C$ is an arbitrary constant)

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $dy=e^{\alpha x+y} dx$; $\alpha \in N$. If $y(\log_{e} 2)=\log_{e} 2$ and $y(0)=\log_{e}(\frac{1}{2})$,then the value of $\alpha$ is equal to $.....$

Let $y=y(x)$ be a solution of the differential equation,$\sqrt{1-x^{2}} \frac{dy}{dx}+\sqrt{1-y^{2}}=0, |x| < 1$. If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$,then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to

Find the general solution of the differential equation: $x^{5} \frac{dy}{dx} = -y^{5}$

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$ is

The general solution of the differential equation $e^{\frac{1}{2}\left(\frac{dy}{dx}\right)}=3^x$ is (where $C$ is a constant of integration).

Vedclass Test Series

Exam Paper Generator

Online Exam Module