The solution of the differential equation fra | vedclass

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

Question

(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv

Vedclass · Accepted Answer

$c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$, where $c$ is an arbitrary constant

Vedclass · Answer

(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.Separating variables: $\int \frac{1 - v}{1 + v^2} dv = \int \frac{1}{x} dx$.$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \ln|x| + C$.$	an^{-1}(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + C = \ln|x| + \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + C$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(x^2 + y^2) - \ln|x| + C = \frac{1}{2} \ln(x^2 + y^2) + C$.$2	an^{-1}(\frac{y}{x}) = \ln(x^2 + y^2) + 2C$.Taking exponential on both sides: $e^{2	an^{-1}(\frac{y}{x})} = e^{\ln(x^2 + y^2)} \cdot e^{2C} = k(x^2 + y^2)$.This matches the form $c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$.

The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$ is

Similar Questions

Identify the statement$(s)$ which is/are True.

Let $y=y(x)$ be the solution of the differential equation $(x^2-3y^2)dx+3xydy=0$ with $y(1)=1$. Then $6y^2(e)$ is equal to $......$

The slope of the tangent at $(x, y)$ to the curve passing through $(2, 1)$ is $\frac{x^2+y^2}{2xy}$. Find the equation of the curve.

Express $\frac{dt}{dx} = \frac{t}{x + t e^{-2x/t}}$ in the form of $\frac{dx}{dt} = \phi\left(\frac{x}{t}\right)$.

If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{2}$,then $P(A \mid B') + P(B \mid A')$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module