If y=xln|cx|,c∈R is the general solution of the DEdydx=yx+ϕxy, then ϕ(2)+ϕ1(2) is

y=xlog|cx|⇒logc+logx=xy⇒0+1x=y⋅1−xy1y2⇒dydx=yx−y2x2⇒ϕ(t)=−1t2⇒ϕ't=2t3∴ϕ(2)+ϕ'2=-14+14=0

If y=xln|cx|,c∈R is the general solution of the DEdydx=yx+ϕxy, then ϕ(2)+ϕ1(2) is

y=xlog|cx|⇒logc+logx=xy⇒0+1x=y⋅1−xy1y2⇒dydx=yx−y2x2⇒ϕ(t)=−1t2⇒ϕ't=2t3∴ϕ(2)+ϕ'2=-14+14=0

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$If y = \frac{x}{l n | cx |}, c \in R is the general solution of the DE \frac{dy}{dx} = \frac{y}{x} + ϕ (\frac{x}{y}), then ϕ (2) + ϕ^{1} (2)$ is

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Detailed Solution

$\begin{array}{l} y = \frac{x}{\log | c x |} \\ \Rightarrow \log c + \log x = \frac{x}{y} \\ \Rightarrow 0 + \frac{1}{x} = \frac{y \cdot 1 - x y^{1}}{y^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} - \frac{y^{2}}{x^{2}} \\ \Rightarrow ϕ (t) = - \frac{1}{t^{2}} \\ \Rightarrow ϕ' (t) = \frac{2}{t^{3}} \\ ∴ ϕ (2) + ϕ' (2) = - \frac{1}{4} + \frac{1}{4} = 0 \end{array}$