Q. \[ \frac{dy}{dx}=\frac{x}{y}. \]

Q: How do I solve \frac{dy}{dx}=\frac{x}{y} by separation of variables?

Rearrange: y\,dy=x\,dx. Integrate: \int y\,dy=\int x\,dx. So \frac{y^2}{2}=\frac{x^2}{2}+C, hence y^2=x^2+C' or y=\pm\sqrt{x^2+C'}.

Q: What is the implicit general solution?

Integrating gives y^2=x^2+C. This implicit form is valid for any real constant C. Differentiating y^2-x^2=C returns 2y\,y'-2x=0, so y'=\frac{x}{y} where y\neq 0.

Q: Is y=0 a solution?

The differential equation has \frac{x}{y} , so y=0 makes the RHS undefined. The only way y=0 could work is if the derivative were undefined as well, which is not allowed for standard solutions. So y=0 is not a (regular) solution.

Q: How do initial conditions work, like y(x_0)=y_0?

Substitute into y^2=x^2+C: y_0^2=x_0^2+C, so C=y_0^2-x_0^2. Then y^2=x^2+y_0^2-x_0^2, and y is chosen as + or - to satisfy the sign at x_0 (if y_0\neq 0).

Q: What happens at points where y changes sign?

Since the equation implies y\,y' = x, solutions satisfy y^2=x^2+C. If y crosses 0, then y^2=0 means x^2=-C. Near such a point, y' blows up because y is in the denominator, so crossing is problematic for differentiable solutions.

Q: Can I write the solution as a family of curves in the xy-plane?

Yes. From y^2=x^2+C, equivalently y^2-x^2=C. Each constant C gives a hyperbola-type curve (or degenerate cases), with vertical and horizontal branches depending on C and the sign choice for y.

Answer

We solve the differential equation \( \frac{dy}{dx}=\frac{x}{y}\) by separating variables:

\[
y\,dy = x\,dx
\]

Integrate both sides:

\[
\int y\,dy = \int x\,dx
\]
\[
\frac{y^2}{2}=\frac{x^2}{2}+C
\]

Multiply by \(2\) and simplify:

\[
y^2=x^2+C
\]

Final result: \(\;y^2=x^2+C\).

Detailed Explanation

We are given the differential equation

\[
\frac{dy}{dx}=\frac{x}{y}.
\]

Step 1: Rewrite the equation in differential form.

Multiply both sides by \(y\) to get all terms involving \(y\) together.

\[
y\,\frac{dy}{dx}=x.
\]

Step 2: Separate variables.

Multiply both sides by \(dx\) and divide by appropriate factors to separate \(y\) and \(x\).

\[
y\,dy = x\,dx.
\]

Step 3: Integrate both sides.

Integrate the left side with respect to \(y\), and the right side with respect to \(x\).

\[
\int y\,dy = \int x\,dx.
\]

Compute each integral.

For the left side:

\[
\int y\,dy = \frac{y^2}{2}.
\]

For the right side:

\[
\int x\,dx = \frac{x^2}{2}.
\]

So the integrated equation is:

\[
\frac{y^2}{2} = \frac{x^2}{2} + C.
\]

Step 4: Simplify the result.

Multiply everything by \(2\) to remove the fractions.

\[
y^2 = x^2 + C_1,
\]

where \(C_1 = 2C\) is another constant.

Equivalently, you can write the constant in the form \(C\) directly:

\[
y^2 = x^2 + C.
\]

Step 5: (Optional) Solve explicitly for \(y\).

If you want \(y\) rather than \(y^2\), take square roots:

\[
y = \pm \sqrt{x^2 + C}.
\]

Final Answer:

\[
y^2 = x^2 + C
\]

or, equivalently,

\[
y = \pm \sqrt{x^2 + C}.
\]

See full solution

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Calculus FAQ

How do I solve \( \frac{dy}{dx}=\frac{x}{y} \) by separation of variables?

Rearrange: \(y\,dy=x\,dx\). Integrate: \(\int y\,dy=\int x\,dx\). So \( \frac{y^2}{2}=\frac{x^2}{2}+C\), hence \(y^2=x^2+C'\) or \(y=\pm\sqrt{x^2+C'}\).

What is the implicit general solution?

Integrating gives \(y^2=x^2+C\). This implicit form is valid for any real constant \(C\). Differentiating \(y^2-x^2=C\) returns \(2y\,y'-2x=0\), so \(y'=\frac{x}{y}\) where \(y\neq 0\).

Is \(y=0\) a solution?

The differential equation has \( \frac{x}{y} \), so \(y=0\) makes the RHS undefined. The only way \(y=0\) could work is if the derivative were undefined as well, which is not allowed for standard solutions. So \(y=0\) is not a (regular) solution.

How do initial conditions work, like \(y(x_0)=y_0\)?

Substitute into \(y^2=x^2+C\): \(y_0^2=x_0^2+C\), so \(C=y_0^2-x_0^2\). Then \(y^2=x^2+y_0^2-x_0^2\), and \(y\) is chosen as \(+\) or \(-\) to satisfy the sign at \(x_0\) (if \(y_0\neq 0\)).

What happens at points where \(y\) changes sign?

Since the equation implies \(y\,y' = x\), solutions satisfy \(y^2=x^2+C\). If \(y\) crosses 0, then \(y^2=0\) means \(x^2=-C\). Near such a point, \(y'\) blows up because \(y\) is in the denominator, so crossing is problematic for differentiable solutions.

Can I write the solution as a family of curves in the \(xy\)-plane?

Yes. From \(y^2=x^2+C\), equivalently \(y^2-x^2=C\). Each constant \(C\) gives a hyperbola-type curve (or degenerate cases), with vertical and horizontal branches depending on \(C\) and the sign choice for \(y\).

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