Separable Equations with Power-Law Dependence

Why the Exponent Changes the Method

The equation \( \frac{dy}{dx}=xy^n \) is separable: its right-hand side is a product of a function of \(x\) and a power of \(y\). Separating variables turns the differential equation into an integration problem. The exponent \(n\) determines an important distinction: when \(n=1\), the \(y\)-side integrates to a logarithm; when \(n\ne 1\), it integrates to a power.

An initial condition \(y(x_0)=y_0\) determines a particular solution, but it does not remove the need to check the solution's domain. In particular, formulas involving noninteger powers of \(y\) are naturally considered on intervals where \(y>0\).

Separating the Variables

Suppose \(y>0\) on the interval being considered. Starting from \( \frac{dy}{dx}=xy^n \), divide by \(y^n\) and multiply by \(dx\):

This form indicates that we integrate the left side with respect to \(y\) and the right side with respect to \(x\). The antiderivative of \(y^{-n}\) depends on whether its exponent is \(-1\), which occurs exactly when \(n=1\).

The Case \(n=1\): Exponential Form

When \(n=1\), the equation is \( \frac{dy}{dx}=xy \). For a nonzero solution, divide by \(y\) and integrate:

Equivalently, on a positive-valued interval, exponentiating gives \(y=Ae^{x^2/2}\) for a positive constant \(A\). If \(y(x_0)=y_0>0\), it is convenient to use the initial value directly:

Indeed, setting \(x=x_0\) gives \(y(x_0)=y_0e^0=y_0\). The same expression works for negative \(y_0\), because the equation is linear in \(y\) and the exponential factor is positive. If \(y_0=0\), the solution is \(y(x)=0\).

The Case \(n\ne 1\): Power Form

For \(n\ne 1\), the power rule for integration gives

Consequently, separating and integrating yields

The initial condition \(y(x_0)=y_0>0\) determines the constant. Subtracting the equation at \(x=x_0\) from the general equation gives

Solving for \(y\) produces the positive-valued solution formula:

The positive bracket condition is not an optional technical detail. A real power with a noninteger exponent need not be defined for a negative base. Also, the derivation divided by \(y^n\), so it applies only where \(y\ne 0\). If \(n>0\), check \(y=0\) separately in the original equation.

Why the Formula Works

The power formula can be justified directly using the chain rule. This also clarifies why the exponent \(1-n\) appears. Assume \(y>0\), \(n\ne1\), and \(y'=xy^n\). Differentiate \(y^{1-n}\):

Integrating with respect to \(x\) gives

If \(y(x_0)=y_0\), substitute \(x_0\) and \(y_0\) to determine \(C_1\):

Substituting this constant back yields

which gives the stated solution after raising both sides to the power \(1/(1-n)\), on an interval where the right-hand side is positive. This derivation shows that the formula is not a pattern to memorize without conditions: it follows from the chain rule and integration, with positivity ensuring the real power is valid.

A Common Integration Mistake

A frequent error is to forget the denominator in the power rule. For \(n=2\), one might incorrectly write

This is wrong: differentiating \(y^{-1}\) gives \(-y^{-2}\), not \(y^{-2}\). The correct integral is

The error changes the resulting solution. For example, using the incorrect integral for \(y'=xy^2\) with \(y(0)=1\) leads to \(1/y=1+x^2/2\), or \(y=1/(1+x^2/2)\). Differentiate that incorrect candidate:

These expressions are not equal in general. The corrected integration gives \( -1/y=x^2/2+C \). Applying \(y(0)=1\) gives \(C=-1\), so \(1/y=1-x^2/2\), matching the checked solution above.

Using Initial Conditions Responsibly

For an initial-value problem, the initial point must lie in the interval where the solution formula is defined. For \(n\ne1\) and \(y_0>0\), the bracket equals \(y_0^{1-n}>0\) at \(x=x_0\), so the initial point is valid. Moving away from it, however, the bracket may reach zero. The positive-valued formula cannot be continued through a point where the bracket is nonpositive.

A reliable workflow is:

These checks distinguish a formal expression from a valid solution to an initial-value problem. The equation, the initial value, and the domain must all agree.