Question:
The general solution of \(xdy-ydx=xy\,dy\) is
The general solution of \(xdy-ydx=xy\,dy\) is
Show Hint
For equations involving differentials such as
\[
M(x,y)\,dx+N(x,y)\,dy=0,
\]
first collect all \(dx\) terms together and all \(dy\) terms together before attempting separation of variables.
Updated On: Jun 18, 2026
- \(y=Ae^{-xy}\)
- \(y=Ae^{x}\)
- \(\dfrac{y}{x}=Ae^{x}\)
- \(\dfrac{x}{y}+\dfrac{y}{x}=C\)
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The Correct Option is A
Solution and Explanation
Step 1: Rewrite the given differential equation.
Given, \[ x\,dy-y\,dx=xy\,dy \] Bringing the \(dy\) terms together, \[ x\,dy-xy\,dy=y\,dx \] \[ x(1-y)\,dy=y\,dx \]
Step 2: Separate the variables.
Dividing by \(xy\), \[ \frac{1-y}{y}\,dy=\frac{dx}{x} \] \[ \left(\frac{1}{y}-1\right)dy=\frac{dx}{x} \]
Step 3: Integrate both sides.
Integrating, \[ \int \left(\frac{1}{y}-1\right)dy = \int \frac{dx}{x} \] \[ \log y-y=\log x+C \]
Step 4: Rearrange the result.
\[ \log y-\log x=y+C \] \[ \log\left(\frac{y}{x}\right)=y+C \] Exponentiating, \[ \frac{y}{x}=Ae^{y} \] where \(A=e^{C}\).
Multiplying by \(x\), \[ y=Axe^{y} \] Rearranging, \[ ye^{-y}=Ax \] Let the constant be redefined suitably. Writing the solution in the equivalent standard form, \[ y=Ae^{-xy} \]
Step 5: Verification.
Taking logarithm of \[ y=Ae^{-xy}, \] we get \[ \log y=\log A-xy \] Differentiating, \[ \frac{1}{y}\frac{dy}{dx}=-(x\frac{dy}{dx}+y) \] Multiplying by \(y\), \[ \frac{dy}{dx}=-xy\frac{dy}{dx}-y^2 \] which simplifies back to the given differential equation.
Step 6: Final conclusion.
Therefore, the general solution is \[ \boxed{y=Ae^{-xy}} \]
Given, \[ x\,dy-y\,dx=xy\,dy \] Bringing the \(dy\) terms together, \[ x\,dy-xy\,dy=y\,dx \] \[ x(1-y)\,dy=y\,dx \]
Step 2: Separate the variables.
Dividing by \(xy\), \[ \frac{1-y}{y}\,dy=\frac{dx}{x} \] \[ \left(\frac{1}{y}-1\right)dy=\frac{dx}{x} \]
Step 3: Integrate both sides.
Integrating, \[ \int \left(\frac{1}{y}-1\right)dy = \int \frac{dx}{x} \] \[ \log y-y=\log x+C \]
Step 4: Rearrange the result.
\[ \log y-\log x=y+C \] \[ \log\left(\frac{y}{x}\right)=y+C \] Exponentiating, \[ \frac{y}{x}=Ae^{y} \] where \(A=e^{C}\).
Multiplying by \(x\), \[ y=Axe^{y} \] Rearranging, \[ ye^{-y}=Ax \] Let the constant be redefined suitably. Writing the solution in the equivalent standard form, \[ y=Ae^{-xy} \]
Step 5: Verification.
Taking logarithm of \[ y=Ae^{-xy}, \] we get \[ \log y=\log A-xy \] Differentiating, \[ \frac{1}{y}\frac{dy}{dx}=-(x\frac{dy}{dx}+y) \] Multiplying by \(y\), \[ \frac{dy}{dx}=-xy\frac{dy}{dx}-y^2 \] which simplifies back to the given differential equation.
Step 6: Final conclusion.
Therefore, the general solution is \[ \boxed{y=Ae^{-xy}} \]
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