Question:
The solution of the differential equation \(\frac{dy}{dx}=\frac{x}{y}\) represents a family of
The solution of the differential equation \(\frac{dy}{dx}=\frac{x}{y}\) represents a family of
Show Hint
Separate the variables to get \(y\,dy=x\,dx\) and integrate.
Updated On: Oct 1, 2026
- circles
- ellipses
- parabolas
- hyperbolas
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We must solve the differential equation and then identify the curve given by its general solution.
Step 2: Key Formula or Approach:
The equation has variables that can be separated. Put all \(y\) terms with \(dy\) and all \(x\) terms with \(dx\), then integrate.
Step 3: Separate and integrate.
\[ y\,dy=x\,dx \]
\[ \frac{y^{2}}{2}=\frac{x^{2}}{2}+k \]
Multiply by 2 and let \(c=2k\):
\[ y^{2}-x^{2}=c \]
Step 4: Identify the curve.
The equation \(x^{2}-y^{2}=-c\) has the form \(x^{2}-y^{2}=\text{constant}\). Such an equation is a hyperbola (a rectangular hyperbola). For \(c=0\) it degenerates into the pair of lines \(y=\pm x\).
Step 5: Check the other options.
A circle or ellipse has a plus sign between \(x^{2}\) and \(y^{2}\), but we have a minus sign. A parabola has only one squared variable. So these three are wrong.
Final Answer:
The solution is a family of hyperbolas, option 4. \[ \boxed{\text{hyperbolas}} \]
We must solve the differential equation and then identify the curve given by its general solution.
Step 2: Key Formula or Approach:
The equation has variables that can be separated. Put all \(y\) terms with \(dy\) and all \(x\) terms with \(dx\), then integrate.
Step 3: Separate and integrate.
\[ y\,dy=x\,dx \]
\[ \frac{y^{2}}{2}=\frac{x^{2}}{2}+k \]
Multiply by 2 and let \(c=2k\):
\[ y^{2}-x^{2}=c \]
Step 4: Identify the curve.
The equation \(x^{2}-y^{2}=-c\) has the form \(x^{2}-y^{2}=\text{constant}\). Such an equation is a hyperbola (a rectangular hyperbola). For \(c=0\) it degenerates into the pair of lines \(y=\pm x\).
Step 5: Check the other options.
A circle or ellipse has a plus sign between \(x^{2}\) and \(y^{2}\), but we have a minus sign. A parabola has only one squared variable. So these three are wrong.
Final Answer:
The solution is a family of hyperbolas, option 4. \[ \boxed{\text{hyperbolas}} \]
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