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(B) Let $u = x - y$. Then $\frac{du}{dx} = 1 - \frac{dy}{dx}$.Substituting $\frac{dy}{dx} = u^2$,we get $1 - \frac{du}{dx} = u^2$,which implies $\frac

Vedclass · Accepted Answer

$- \log_e \left| \frac{1 - x + y}{1 + x - y} \right| = 2(x - 1)$

Vedclass · Answer

(B) Let $u = x - y$. Then $\frac{du}{dx} = 1 - \frac{dy}{dx}$.Substituting $\frac{dy}{dx} = u^2$,we get $1 - \frac{du}{dx} = u^2$,which implies $\frac{du}{dx} = 1 - u^2$.Separating the variables,we have $\frac{du}{1 - u^2} = dx$.Integrating both sides,$\int \frac{du}{1 - u^2} = \int dx$,which gives $\frac{1}{2} \log_e \left| \frac{1 + u}{1 - u} \right| = x + C$.Substituting $u = x - y$,we get $\frac{1}{2} \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = x + C$.Using the condition $y(1) = 1$,we have $x = 1, y = 1$,so $u = 1 - 1 = 0$.$\frac{1}{2} \log_e \left| \frac{1 + 0}{1 - 0} \right| = 1 + C \Rightarrow 0 = 1 + C \Rightarrow C = -1$.Thus,$\frac{1}{2} \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = x - 1$,which simplifies to $\log_e \left| \frac{1 + x - y}{1 - x + y} \right| = 2(x - 1)$.Multiplying by $-1$ on both sides,we get $- \log_e \left| \frac{1 + x - y}{1 - x + y} \right| = -2(x - 1)$,which is equivalent to $- \log_e \left| \frac{1 - x + y}{1 + x - y} \right| = 2(x - 1)$.

The solution of the differential equation $\frac{dy}{dx} = (x - y)^2$ when $y(1) = 1$ is:

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