The solution of x2 y-x3 dydx=y4 cos x, given that x=π, y=π is

x2y-x3dydx=y4 cosx ⇒3dydxy4+3xy3=3 cos xx3Let V=1y3⇒dvdx=-3dydxy4∴dvdx+3vx=3cos xx3Let μ(x)=e∫3 dx=x3multiple both sides by μ(x)x3dvdx+(3x2)V=3 cosx ⇒x3dvdx+ddx(x3)V=3 cosxApplying gdfdx+fdgdx=ddx(fg)ddx(x3v)=3 cosx ⇒∫ddx(x3v)dx=∫3 cosx dx ⇒∫x3V=3 sin x+c ⇒x3 y-3=3 sin x+c ⇒x3y3=3 sin x+c ⇒x3=y3(1+3 sinx)

The solution of x2 y-x3 dydx=y4 cos x, given that x=π, y=π is

x2y-x3dydx=y4 cosx ⇒3dydxy4+3xy3=3 cos xx3Let V=1y3⇒dvdx=-3dydxy4∴dvdx+3vx=3cos xx3Let μ(x)=e∫3 dx=x3multiple both sides by μ(x)x3dvdx+(3x2)V=3 cosx ⇒x3dvdx+ddx(x3)V=3 cosxApplying gdfdx+fdgdx=ddx(fg)ddx(x3v)=3 cosx ⇒∫ddx(x3v)dx=∫3 cosx dx ⇒∫x3V=3 sin x+c ⇒x3 y-3=3 sin x+c ⇒x3y3=3 sin x+c ⇒x3=y3(1+3 sinx)

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The solution of $x^{2} y - x^{3} \frac{dy}{dx} = y^{4} \cos x,$ given that $x = π, y = π$ is

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$y^{3} (1 + 3 \sin x) = x^{3}$

$x^{3} (1 + \sin x) = y^{3}$

$y^{3} (1 - \sin x) = x^{3}$

$x^{3} (1 - \sin x) = y^{3}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} x^{2} y - x^{3} \frac{dy}{dx} & = & y^{4} cosx \\ \Rightarrow & \frac{3 \frac{dy}{dx}}{y^{4}} + \frac{3}{{xy}^{3}} = \frac{3 \cos x}{x^{3}} \\ Let V & = & \frac{1}{y^{3}} \Rightarrow \frac{dv}{dx} = \frac{- 3 \frac{dy}{dx}}{y^{4}} \\ ∴ \frac{dv}{dx} + \frac{3 v}{x} & = & \frac{3 \cos x}{x^{3}} \\ Let μ (x) & = & e^{\int 3 dx} = x^{3} \end{array}$

multiple both sides by $μ (x)$

$\begin{array}{rcl} x^{3} \frac{dv}{dx} + (3 x^{2}) V & = & 3 cosx \\ \Rightarrow & x^{3} \frac{dv}{dx} + \frac{d}{dx} (x^{3}) V = 3 cosx \\ Applying g \frac{df}{dx} + f \frac{dg}{dx} & = & \frac{d}{dx} (fg) \\ \frac{d}{dx} (x^{3} v) & = & 3 cosx \\ \Rightarrow & \int \frac{d}{dx} (x^{3} v) dx = \int 3 cosx dx \\ \Rightarrow & \int x^{3} V = 3 \sin x + c \\ \Rightarrow & x^{3} y^{- 3} = 3 \sin x + c \\ \Rightarrow & \frac{x^{3}}{y^{3}} = 3 \sin x + c \\ \Rightarrow & x^{3} = y^{3} (1 + 3 sinx) \end{array}$