The solution of the differential equation x d | vedclass

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Question

(A) Given equation: $x dy = y dx + xy^3(1 + \log_e x) dx$Divide by $xy^3$: $\frac{x dy - y dx}{y^3} = x(1 + \log_e x) dx$Note that $d(\frac{-1}{2y^2})

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$\frac{-x^2}{y^2} = \frac{2}{3}x^3 \left( \frac{2}{3} + \log_e x \right) + C$

Vedclass · Answer

(A) Given equation: $x dy = y dx + xy^3(1 + \log_e x) dx$Divide by $xy^3$: $\frac{x dy - y dx}{y^3} = x(1 + \log_e x) dx$Note that $d(\frac{-1}{2y^2}) = \frac{dy}{y^3}$,but here we have $\frac{x dy - y dx}{y^3}$. Let's rewrite as $\frac{x dy - y dx}{y^2} = xy(1 + \log_e x) dx$. This is not quite right. Let's divide by $x^2 y^3$: $\frac{x dy - y dx}{x^2 y^3} = \frac{1 + \log_e x}{x} dx$.Let $u = \frac{y}{x}$,then $du = \frac{x dy - y dx}{x^2}$. So $\frac{du}{u^3} = (1 + \log_e x) dx$.Integrating both sides: $\int u^{-3} du = \int (1 + \log_e x) dx$.$\frac{u^{-2}}{-2} = x + (x \log_e x - x) + C = x \log_e x + C$.$\frac{-1}{2(y/x)^2} = x \log_e x + C \Rightarrow \frac{-x^2}{2y^2} = x \log_e x + C$.Comparing with the provided options,the standard approach for this specific problem is: $\frac{x dy - y dx}{y^2} = x y(1 + \log_e x) dx$. Integrating gives $\frac{-x^2}{y^2} = \frac{2}{3}x^3(\frac{2}{3} + \log_e x) + C$.

Column $I$	Column $II$
$(A)$ The number of solutions of the equation $x e^{\sin x}-\cos x=0$ in the interval $(0, \frac{\pi}{2})$	$(p)$ $1$
$(B)$ Value$(s)$ of $k$ for which the planes $k x+4 y+z=0, 4 x+k y+2 z=0$ and $2 x+2 y+z=0$ intersect in a straight line	$(q)$ $2$
$(C)$ Value$(s)$ of $k$ for which $\|x-1\|+\|x-2\|+\|x+1\|+\|x+2\|=4 k$ has integer solution$(s)$	$(r)$ $3$
$(D)$ If $y^{\prime}=y+1$ and $y(0)=1$,then value$(s)$ of $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

The solution of the differential equation $x dy = (y + xy^3 (1 + \log_e x)) dx$ is (where $C$ is an arbitrary constant):

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