If y frac{dy}{dx} = x left[ frac{y^2}{x^2} + | vedclass

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(B) Let $v = \frac{y^2}{x^2}$,so $y^2 = v x^2$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2vx^2 + x^2 \frac{dv}{dx}$,which implie

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$4 \phi(1)$

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(B) Let $v = \frac{y^2}{x^2}$,so $y^2 = v x^2$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2vx^2 + x^2 \frac{dv}{dx}$,which implies $y \frac{dy}{dx} = vx^2 + \frac{x^2}{2} \frac{dv}{dx}$.Substituting this into the given equation: $vx^2 + \frac{x^2}{2} \frac{dv}{dx} = x \left[ v + \frac{\phi(v)}{\phi'(v)} \right] = xv + x \frac{\phi(v)}{\phi'(v)}$.Since $x > 0$,we divide by $x$: $vx + \frac{x}{2} \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.This simplifies to $\frac{x}{2} \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)} + v(1-x)$. This approach is complex,so let's use $y^2 = u$. Then $2y dy = du$,so $y dy = \frac{1}{2} du$.The equation becomes $\frac{1}{2} \frac{du}{dx} = x \left[ \frac{u}{x^2} + \frac{\phi(u/x^2)}{\phi'(u/x^2)} \right] = \frac{u}{x} + x \frac{\phi(u/x^2)}{\phi'(u/x^2)}$.Let $u = v x^2$,then $\frac{du}{dx} = v(2x) + x^2 \frac{dv}{dx}$.Substituting: $\frac{1}{2} (2vx + x^2 \frac{dv}{dx}) = vx + x \frac{\phi(v)}{\phi'(v)}$.$vx + \frac{x^2}{2} \frac{dv}{dx} = vx + x \frac{\phi(v)}{\phi'(v)}$.$\frac{x}{2} \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)} \implies \frac{\phi'(v)}{\phi(v)} dv = \frac{2}{x} dx$.Integrating both sides: $\ln(\phi(v)) = 2 \ln(x) + C = \ln(x^2) + C$.So,$\phi(v) = k x^2$,where $k = e^C$.Since $v = y^2/x^2$,we have $\phi(y^2/x^2) = k x^2$.Given $y(1) = -1$,at $x=1$,$v = (-1)^2/1^2 = 1$. Thus $\phi(1) = k(1)^2 = k$.We want to find $\phi(y^2/4)$. Since $v = y^2/x^2$,if we set $x=2$,then $v = y^2/4$.Therefore,$\phi(y^2/4) = k(2)^2 = 4k = 4 \phi(1)$.

If $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$,$x > 0$,$\phi > 0$,and $y(1) = -1$,then $\phi(y^2/4)$ is equal to:

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