The general solution of the differential equa | vedclass

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Question

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

Vedclass · Accepted Answer

$y = \log_e\left(\frac{x+y+2}{k}\right)$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{1}{x+y+1}$.Let $v = x+y+1$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting this into the equation: $\frac{dv}{dx} - 1 = \frac{1}{v}$.$\frac{dv}{dx} = \frac{1}{v} + 1 = \frac{1+v}{v}$.Separating the variables: $\frac{v}{1+v} dv = dx$.$\left(\frac{1+v-1}{1+v}\right) dv = dx \Rightarrow \left(1 - \frac{1}{1+v}\right) dv = dx$.Integrating both sides: $\int (1 - \frac{1}{1+v}) dv = \int dx$.$v - \log|1+v| = x + c$.Substitute $v = x+y+1$: $(x+y+1) - \log|x+y+2| = x + c$.$y + 1 - \log|x+y+2| = c$.$y = \log|x+y+2| + c - 1$.Let $c - 1 = \log k$, then $y = \log|x+y+2| - \log k = \log\left(\frac{x+y+2}{k}\right)$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{1}{x+y+1}$ is ($k, c$ are arbitrary constants)

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