A curve passes through the point left( 1, fra | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Le

Vedclass · Accepted Answer

$y = x 	an^{-1} \left( \ln \frac{e}{x} ight)$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} ight)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v - \cos^2 v$.Simplifying,we get: $x \frac{dv}{dx} = -\cos^2 v$.Separating the variables: $\frac{dv}{\cos^2 v} = -\frac{dx}{x}$,which is $\sec^2 v \, dv = -\frac{dx}{x}$.Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{1}{x} \, dx$.This gives: $	an v = -\ln |x| + C$.Substituting $v = \frac{y}{x}$: $	an \left( \frac{y}{x} ight) = -\ln x + C$.The curve passes through $(1, \frac{\pi}{4})$,so $	an \left( \frac{\pi/4}{1} ight) = -\ln(1) + C$.Since $	an(\frac{\pi}{4}) = 1$ and $\ln(1) = 0$,we have $1 = 0 + C$,so $C = 1$.Thus,$	an \left( \frac{y}{x} ight) = 1 - \ln x = \ln e - \ln x = \ln \left( \frac{e}{x} ight)$.Therefore,$y = x 	an^{-1} \left( \ln \frac{e}{x} ight)$.

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Then the equation of the curve is:

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