Question:
The general solution of the differential equation \( y\,dx-x\,dy=y^2(x\,dy+y\,dx) \) is
The general solution of the differential equation \( y\,dx-x\,dy=y^2(x\,dy+y\,dx) \) is
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In differential equations involving \( dx \) and \( dy \), first collect like terms carefully. After separation, partial fractions often make the integration straightforward.
Updated On: Apr 28, 2026
- \( \dfrac{y}{x}=xy+C \)
- \( \dfrac{x}{y^2}=xy+C \)
- \( \dfrac{y}{x^2}=xy+C \)
- \( x+y=xy+C \)
- \( \dfrac{x}{y}=xy+C \)
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The Correct Option is
Solution and Explanation
Step 1: Write the given differential equation clearly.
We are given \[ y\,dx-x\,dy=y^2(x\,dy+y\,dx) \] We need to simplify this equation and then integrate it to get the general solution.
Step 2: Expand the right-hand side.
Distribute \( y^2 \) on the right: \[ y\,dx-x\,dy=x y^2\,dy+y^3\,dx \] Now bring the terms involving \( dx \) together and the terms involving \( dy \) together.
Step 3: Rearrange the equation.
Move all terms to one side: \[ y\,dx-y^3\,dx=x\,dy+x y^2\,dy \] Factor both sides: \[ y(1-y^2)\,dx=x(1+y^2)\,dy \] Now divide both sides by \( xy \): \[ \frac{1-y^2}{x}\,dx=\frac{1+y^2}{y}\,dy \] A better rearrangement is: \[ \frac{dx}{x}=\frac{1+y^2}{y(1-y^2)}\,dy \]
Step 4: Split the right-hand side into partial fractions.
We simplify \[ \frac{1+y^2}{y(1-y^2)} \] Using partial fractions: \[ \frac{1+y^2}{y(1-y^2)}=\frac{1}{y}+\frac{2y}{1-y^2} \] So the differential equation becomes \[ \frac{dx}{x}=\left(\frac{1}{y}+\frac{2y}{1-y^2}\right)dy \]
Step 5: Integrate both sides.
Integrating, \[ \int \frac{dx}{x}=\int \left(\frac{1}{y}+\frac{2y}{1-y^2}\right)dy \] Now, \[ \int \frac{dx}{x}=\ln|x| \] Also, \[ \int \frac{1}{y}\,dy=\ln|y| \] and for \[ \int \frac{2y}{1-y^2}\,dy \] let \[ u=1-y^2 \Rightarrow du=-2y\,dy \] So, \[ \int \frac{2y}{1-y^2}\,dy=-\ln|1-y^2| \] Hence, \[ \ln|x|=\ln|y|-\ln|1-y^2|+C \]
Step 6: Simplify the logarithmic result.
Using log laws, \[ \ln|x|=\ln\left|\frac{y}{1-y^2}\right|+C \] So, \[ x=C\cdot \frac{y}{1-y^2} \] Rearranging, \[ x(1-y^2)=Cy \] \[ x-xy^2=Cy \] Now divide by \( y \): \[ \frac{x}{y}-xy=C \] Therefore, \[ \frac{x}{y}=xy+C \]
Step 7: Final conclusion.
Hence, the general solution is \[ \boxed{\frac{x}{y}=xy+C} \] Therefore, the correct option is \[ \boxed{(5)\ \dfrac{x}{y}=xy+C} \]
We are given \[ y\,dx-x\,dy=y^2(x\,dy+y\,dx) \] We need to simplify this equation and then integrate it to get the general solution.
Step 2: Expand the right-hand side.
Distribute \( y^2 \) on the right: \[ y\,dx-x\,dy=x y^2\,dy+y^3\,dx \] Now bring the terms involving \( dx \) together and the terms involving \( dy \) together.
Step 3: Rearrange the equation.
Move all terms to one side: \[ y\,dx-y^3\,dx=x\,dy+x y^2\,dy \] Factor both sides: \[ y(1-y^2)\,dx=x(1+y^2)\,dy \] Now divide both sides by \( xy \): \[ \frac{1-y^2}{x}\,dx=\frac{1+y^2}{y}\,dy \] A better rearrangement is: \[ \frac{dx}{x}=\frac{1+y^2}{y(1-y^2)}\,dy \]
Step 4: Split the right-hand side into partial fractions.
We simplify \[ \frac{1+y^2}{y(1-y^2)} \] Using partial fractions: \[ \frac{1+y^2}{y(1-y^2)}=\frac{1}{y}+\frac{2y}{1-y^2} \] So the differential equation becomes \[ \frac{dx}{x}=\left(\frac{1}{y}+\frac{2y}{1-y^2}\right)dy \]
Step 5: Integrate both sides.
Integrating, \[ \int \frac{dx}{x}=\int \left(\frac{1}{y}+\frac{2y}{1-y^2}\right)dy \] Now, \[ \int \frac{dx}{x}=\ln|x| \] Also, \[ \int \frac{1}{y}\,dy=\ln|y| \] and for \[ \int \frac{2y}{1-y^2}\,dy \] let \[ u=1-y^2 \Rightarrow du=-2y\,dy \] So, \[ \int \frac{2y}{1-y^2}\,dy=-\ln|1-y^2| \] Hence, \[ \ln|x|=\ln|y|-\ln|1-y^2|+C \]
Step 6: Simplify the logarithmic result.
Using log laws, \[ \ln|x|=\ln\left|\frac{y}{1-y^2}\right|+C \] So, \[ x=C\cdot \frac{y}{1-y^2} \] Rearranging, \[ x(1-y^2)=Cy \] \[ x-xy^2=Cy \] Now divide by \( y \): \[ \frac{x}{y}-xy=C \] Therefore, \[ \frac{x}{y}=xy+C \]
Step 7: Final conclusion.
Hence, the general solution is \[ \boxed{\frac{x}{y}=xy+C} \] Therefore, the correct option is \[ \boxed{(5)\ \dfrac{x}{y}=xy+C} \]
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