If the solution of frac{dy}{dx} = frac{y^3 co | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}$.Separating the variables,we get $\int \frac{e^{1/y^

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$e - 4 \sin \sqrt{x}$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}$.Separating the variables,we get $\int \frac{e^{1/y^2}}{y^3} dy = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$.Let $t = \frac{1}{y^2}$,then $dt = -\frac{2}{y^3} dy$,so $\frac{dy}{y^3} = -\frac{dt}{2}$.Let $u = \sqrt{x}$,then $du = \frac{dx}{2\sqrt{x}}$,so $\frac{dx}{\sqrt{x}} = 2du$.Substituting these into the integral,we get $\int e^t (-\frac{1}{2}) dt = \int \cos u (2 du)$.$-\frac{1}{2} e^t = 2 \sin u + C$.Substituting back $t = \frac{1}{y^2}$ and $u = \sqrt{x}$,we have $-\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} + C$.Given $y(0) = 1$,at $x = 0$,$y = 1$,so $-\frac{1}{2} e^1 = 2 \sin(0) + C$,which gives $C = -\frac{e}{2}$.Thus,$-\frac{1}{2} e^{1/y^2} = 2 \sin \sqrt{x} - \frac{e}{2}$.Multiplying by $-2$,we get $e^{1/y^2} = e - 4 \sin \sqrt{x}$.Taking the natural logarithm on both sides,$\frac{1}{y^2} = \log_e(e - 4 \sin \sqrt{x})$.Comparing this with $\frac{1}{y^2} = \log_e(f(x))$,we get $f(x) = e - 4 \sin \sqrt{x}$.

If the solution of $\frac{dy}{dx} = \frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1/y^2}}$ with $y(0) = 1$ is $\frac{1}{y^2} = \log_e(f(x))$,then $f(x) =$

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