The general solution of the differential equa | vedclass

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Question

(A) Given differential equation: $(x-y^{2}) dx + y(5x+y^{2}) dy = 0$.Rewrite as: $\frac{dy}{dx} = \frac{y^{2}-x}{y(5x+y^{2})}$.Let $v = y^{2}$,then $\

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$(y^{2}+x)^{4} = C|y^{2}+2x|^{3}$

Vedclass · Answer

(A) Given differential equation: $(x-y^{2}) dx + y(5x+y^{2}) dy = 0$.Rewrite as: $\frac{dy}{dx} = \frac{y^{2}-x}{y(5x+y^{2})}$.Let $v = y^{2}$,then $\frac{dv}{dx} = 2y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2y} \frac{dv}{dx}$.Substituting into the equation: $\frac{1}{2y} \frac{dv}{dx} = \frac{v-x}{y(5x+v)} \implies \frac{dv}{dx} = 2 \frac{v-x}{5x+v}$.Let $v = kx$,then $\frac{dv}{dx} = k + x \frac{dk}{dx}$.$k + x \frac{dk}{dx} = 2 \frac{kx-x}{5x+kx} = 2 \frac{k-1}{5+k}$.$x \frac{dk}{dx} = \frac{2k-2}{k+5} - k = \frac{2k-2-k^{2}-5k}{k+5} = -\frac{k^{2}+3k+2}{k+5} = -\frac{(k+1)(k+2)}{k+5}$.Separating variables: $\int \frac{k+5}{(k+1)(k+2)} dk = -\int \frac{dx}{x}$.Using partial fractions: $\frac{k+5}{(k+1)(k+2)} = \frac{4}{k+1} - \frac{3}{k+2}$.Integrating: $4 \ln|k+1| - 3 \ln|k+2| = -\ln|x| + \ln|C|$.$\ln|\frac{(k+1)^{4}}{(k+2)^{3}}| = \ln|\frac{C}{x}| \implies \frac{(k+1)^{4}}{(k+2)^{3}} = \frac{C}{x}$.Substitute $k = \frac{v}{x} = \frac{y^{2}}{x}$: $\frac{(\frac{y^{2}}{x}+1)^{4}}{(\frac{y^{2}}{x}+2)^{3}} = \frac{C}{x} \implies \frac{(y^{2}+x)^{4}}{x^{4}} \cdot \frac{x^{3}}{(y^{2}+2x)^{3}} = \frac{C}{x}$.$(y^{2}+x)^{4} = C|y^{2}+2x|^{3}$.

The general solution of the differential equation $(x-y^{2}) dx + y(5x+y^{2}) dy = 0$ is:

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