The solution of the differential equation fra | vedclass

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Question

(C) Given the differential equation: $\frac{dy}{dx} + 3 = \frac{y+3x}{\log_{e}(y+3x)}$.Let $z = y + 3x$. Then $\frac{dz}{dx} = \frac{dy}{dx} + 3$.Subs

Vedclass · Accepted Answer

$x - \frac{1}{2}(\log_{e}(y+3x))^{2} = C$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} + 3 = \frac{y+3x}{\log_{e}(y+3x)}$.Let $z = y + 3x$. Then $\frac{dz}{dx} = \frac{dy}{dx} + 3$.Substituting this into the equation,we get $\frac{dz}{dx} = \frac{z}{\log_{e}z}$.Rearranging the terms for variable separation: $\frac{\log_{e}z}{z} dz = dx$.Integrating both sides: $\int \frac{\log_{e}z}{z} dz = \int dx$.Let $u = \log_{e}z$,then $du = \frac{1}{z} dz$. The integral becomes $\int u du = x + C$.So,$\frac{u^{2}}{2} = x + C$.Substituting $u = \log_{e}(y+3x)$ back,we get $\frac{1}{2}(\log_{e}(y+3x))^{2} = x + C$.Rearranging gives $x - \frac{1}{2}(\log_{e}(y+3x))^{2} = C$.

The solution of the differential equation $\frac{dy}{dx} - \frac{y+3x}{\log_{e}(y+3x)} + 3 = 0$ is (where $C$ is a constant of integration.)

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