The solution of the equation frac{dy}{dx} = f | vedclass

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}$.Factorizing the quadratic expressions: $\frac{dy}{dx} = \frac{

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$4\log \left| \frac{y - 2}{y + 1} \right| = 3\log \left| \frac{x - 1}{x + 3} \right| + c$

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}$.Factorizing the quadratic expressions: $\frac{dy}{dx} = \frac{(y - 2)(y + 1)}{(x + 3)(x - 1)}$.Separating the variables: $\frac{dy}{(y - 2)(y + 1)} = \frac{dx}{(x + 3)(x - 1)}$.Integrating both sides: $\int \frac{dy}{(y - 2)(y + 1)} = \int \frac{dx}{(x + 3)(x - 1)}$.Using partial fractions: $\frac{1}{3} \int \left( \frac{1}{y - 2} - \frac{1}{y + 1} \right) dy = \frac{1}{4} \int \left( \frac{1}{x - 1} - \frac{1}{x + 3} \right) dx$.Integrating: $\frac{1}{3} \log \left| \frac{y - 2}{y + 1} \right| = \frac{1}{4} \log \left| \frac{x - 1}{x + 3} \right| + C_1$.Multiplying by $12$: $4 \log \left| \frac{y - 2}{y + 1} \right| = 3 \log \left| \frac{x - 1}{x + 3} \right| + c$.

The solution of the equation $\frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}$ is

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