The general solution of the differential equa | vedclass

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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 $\fr

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

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