The general solution of the differential equa | vedclass

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

Question

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$.Let $v = x+y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\fra

Vedclass · Accepted Answer

$(x+y-5)^2 = C e^{y-x}$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$.Let $v = x+y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting this into the equation: $\frac{dv}{dx} - 1 = \frac{v-3}{v-7}$.$\frac{dv}{dx} = \frac{v-3}{v-7} + 1 = \frac{v-3+v-7}{v-7} = \frac{2v-10}{v-7} = \frac{2(v-5)}{v-7}$.Separating the variables: $\int \frac{v-7}{v-5} dv = \int 2 dx$.$\int \frac{(v-5)-2}{v-5} dv = 2x + C$.$\int (1 - \frac{2}{v-5}) dv = 2x + C$.$v - 2 \ln |v-5| = 2x + C$.Substitute $v = x+y$: $(x+y) - 2 \ln |x+y-5| = 2x + C$.$y-x-C = 2 \ln |x+y-5|$.$\ln |x+y-5|^2 = y-x-C$.Taking the exponential of both sides: $(x+y-5)^2 = e^{y-x-C} = C' e^{y-x}$.Thus,the correct option is $B$.

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

Similar Questions

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

The equation of a curve passing through the origin,if the slope of the tangent drawn at any of its points $(x, y)$ is $\cos (x + y) + \sin (x + y)$,is

The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is

The general solution of ${x^2}\frac{{dy}}{{dx}} = 2$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module