Mixed Separable Initial-Value Problem Workshop

A Reliable Method for Separable Initial-Value Problems

A separable initial-value problem combines a differential equation with a known value of the unknown function at a particular input. The goal is to find a function that satisfies both. The calculations often involve familiar antiderivatives, but a complete solution also requires keeping track of restrictions introduced by division, choosing the correct constant from the initial value, and checking that the resulting function solves the original equation.

We will work through examples with exponential, power-law, trigonometric, and rational factors. In every case, the final check will return to the differential equation itself.

Definition: A differential equation is separable if it can be written as \(y'=g(t)h(y)\), where \(g\) depends only on the independent variable \(t\) and \(h\) depends only on the dependent variable \(y\). An initial-value problem also specifies \(y(t_0)=y_0\). A solution is a differentiable function on an interval containing \(t_0\) that satisfies both the equation and the initial condition.

Separation, Integration, and the Initial Condition

When \(h(y)\ne 0\), the equation \(y'=g(t)h(y)\) can be rearranged to place the \(y\)-terms with \(dy\) and the \(t\)-terms with \(dt\):

\frac{1}{h(y)}\,dy=g(t)\,dt.

Integrating gives a relation between \(y\) and \(t\). The initial condition then determines the constant of integration. Sometimes this relation can be solved explicitly for \(y\); sometimes it is most useful in implicit form.

The restriction \(h(y)\ne 0\) matters: division by \(h(y)\) is not valid where \(h(y)=0\). If \(h(y_*)=0\), then the constant function \(y(t)=y_*\) satisfies the differential equation. Check whether such an equilibrium solution meets the initial condition before dividing.

Why the Separation Method Works

The method follows from the chain rule. Suppose \(H\) is an antiderivative of \(1/h\) on a region where \(h\ne 0\), and \(G\) is an antiderivative of \(g\). If \(y(t)\) solves the differential equation, then:

\frac{d}{dt}H(y(t)) =H'(y(t))y'(t) =\frac{1}{h(y(t))}\,g(t)h(y(t)) =g(t) =G'(t).

Thus \(H(y(t))\) and \(G(t)\) have the same derivative, so they differ by a constant on the interval under consideration. This proves the separated relation \(H(y)=G(t)+C\). Conversely, differentiating that relation where \(H'(y)=1/h(y)\) is defined recovers \(y'=g(t)h(y)\). The initial condition determines \(C\), and any explicit formula must still be restricted to an interval where it is defined and the derivation remains valid.

Worked Examples

Worked Example: An Exponential Factor and a Quadratic Power of \(y\)

Solve \(y'=e^t y^2\), with \(y(0)=1\).

Since the initial value is nonzero, separate variables on an interval where \(y\ne 0\):

\frac{1}{y^2}\,dy=e^t\,dt.

Integrate each side. The antiderivative of \(y^{-2}\) is \(-y^{-1}\), not \(\ln|y|\):

-\frac{1}{y}=e^t+C.

Use \(y(0)=1\) to find \(C\):

-1=e^0+C=1+C, \qquad C=-2.

Therefore \(-1/y=e^t-2\), so \(1/y=2-e^t\) and

y(t)=\frac{1}{2-e^t}.

The denominator is nonzero when \(e^t\ne 2\), or \(t\ne\ln 2\). The interval containing the initial time \(0\) on which the formula is defined is \((-\infty,\ln 2)\).

To verify, differentiate and substitute:

y'(t)=\frac{e^t}{(2-e^t)^2}, \qquad e^t y(t)^2 =e^t\left(\frac{1}{2-e^t}\right)^2 =\frac{e^t}{(2-e^t)^2}.

The two expressions agree, and \(y(0)=1/(2-1)=1\), as required.

Worked Example: A Power-Law Factor

Solve \(y'=2t\sqrt{y}\), with \(y(0)=1\).

On an interval where \(y>0\), separate and integrate:

\frac{1}{\sqrt{y}}\,dy=2t\,dt, \qquad \int y^{-1/2}\,dy=\int 2t\,dt, \qquad 2\sqrt{y}=t^2+C.

Substitute the initial condition:

2\sqrt{1}=0^2+C, \qquad C=2.

Thus \(2\sqrt{y}=t^2+2\), so \(\sqrt{y}=1+t^2/2\). Squaring gives

y(t)=\left(1+\frac{t^2}{2}\right)^2.

For every real \(t\), \(1+t^2/2\ge 1\), so \(y(t)>0\); the square root used in the equation is therefore \(\sqrt{y}=1+t^2/2\). Verify by differentiating and comparing with the right-hand side:

y'(t)=2\left(1+\frac{t^2}{2}\right)t, \qquad 2t\sqrt{y(t)} =2t\left(1+\frac{t^2}{2}\right).

These are equal, and \(y(0)=(1+0)^2=1\). The formula satisfies the initial-value problem for all real \(t\).

Worked Example: A Trigonometric Factor

Solve \(y'=(1+y^2)\cos t\), with \(y(0)=0\).

Here \(1+y^2\) is always positive, so division by it is valid for every real \(y\). Separate and integrate:

\frac{1}{1+y^2}\,dy=\cos t\,dt, \qquad \arctan y=\sin t+C.

At \(t=0\), the condition \(y(0)=0\) gives \(\arctan(0)=\sin(0)+C\), hence \(C=0\). Taking the tangent of both sides yields

y(t)=\tan(\sin t).

The tangent is defined provided its input is not an odd multiple of \(\pi/2\). Since \(-1\le\sin t\le 1\), and \(1<\pi/2\), the input always lies strictly between \(-\pi/2\) and \(\pi/2\). Thus the formula is defined for every real \(t\).

Differentiate using the chain rule, then substitute into the equation:

y'(t)=\sec^2(\sin t)\cos t,

(1+y(t)^2)\cos t =\left(1+\tan^2(\sin t)\right)\cos t =\sec^2(\sin t)\cos t.

The expressions agree, and \(y(0)=\tan(0)=0\).

Worked Example: Rational Factors on Both Sides

Solve \(y'=\dfrac{t}{1+t^2}(1-y^2)\), with \(y(0)=\dfrac12\).

For \(y\ne\pm1\), separate variables:

\frac{1}{1-y^2}\,dy=\frac{t}{1+t^2}\,dt.

Use \(1/(1-y^2)=\frac12(1/(1+y)+1/(1-y))\) to integrate the left side. On the branch containing \(y=1/2\), where \(-1<y<1\), this gives:

\frac12\ln\left(\frac{1+y}{1-y}\right) =\frac12\ln(1+t^2)+C.

At \(t=0\), the logarithm on the left is \(\frac12\ln 3\) and the logarithm on the right is \(0\). Hence \(C=\frac12\ln 3\). Multiply by \(2\), exponentiate, and solve for \(y\):

\ln\left(\frac{1+y}{1-y}\right)=\ln\bigl(3(1+t^2)\bigr), \qquad \frac{1+y}{1-y}=3(1+t^2),

y(t)=\frac{3(1+t^2)-1}{3(1+t^2)+1} =\frac{2+3t^2}{4+3t^2}.

The denominator \(4+3t^2\) is positive for every real \(t\), and the numerator is positive and smaller than the denominator. Therefore \(0<y(t)<1\), so the separation restriction \(y\ne\pm1\) holds for all real \(t\). Also \(1+t^2\) never vanishes.

To check the equation, differentiate the quotient and simplify its right-hand side:

y'(t) =\frac{6t(4+3t^2)-6t(2+3t^2)}{(4+3t^2)^2} =\frac{12t}{(4+3t^2)^2},

\frac{t}{1+t^2}(1-y^2) =\frac{t}{1+t^2} \frac{(4+3t^2)^2-(2+3t^2)^2}{(4+3t^2)^2} =\frac{t}{1+t^2} \frac{12(1+t^2)}{(4+3t^2)^2} =\frac{12t}{(4+3t^2)^2}.

Thus the differential equation holds, and \(y(0)=2/4=1/2\).

A Pitfall: Dividing Away a Solution

In the first example, dividing by \(y^2\) required \(y\ne0\). The original equation \(y'=e^t y^2\) also has the constant solution \(y(t)=0\), because both sides equal zero. It does not satisfy the example’s initial condition \(y(0)=1\), but an initial condition \(y(0)=0\) would be met by that equilibrium solution. Dividing by \(y^2\) in that different problem would discard it.

There is also a common integration error in the same separation step. A mistaken calculation might say:

\int y^{-2}\,dy=\ln|y|.

This is incorrect: \(\ln|y|\) differentiates to \(1/y\), not \(1/y^2\). The power rule gives the correct calculation and, in the first example, the corresponding solution:

\int y^{-2}\,dy=-\frac{1}{y}+C, \qquad y(t)=\frac{1}{2-e^t}.

A Final Checklist

For a separable initial-value problem, first identify any values at which division would be invalid and check for equilibrium solutions. Then separate and integrate, apply the initial condition, and solve for \(y\) when possible. Finally, check the formula’s domain and substitute its derivative and value back into the original problem. These checks distinguish a valid solution on the required interval from an algebraic expression that merely resembles one.

Check Your Understanding

For each problem, try to find the solution and state an interval containing the initial time on which it is valid.

For \(y'=3e^t y\), \(y(0)=2\), what solution satisfies the initial condition?
For \(y'=t y^{1/3}\), \(y(0)=1\), what restriction must be considered when separating variables?
In \(y'=(1+y^2)\sin t\), \(y(0)=0\), why is division by \(1+y^2\) valid for every real \(y\)?
For \(y'=t(1-y^2)\), \(y(0)=1\), what solution would be lost by dividing by \(1-y^2\)?
What two checks should follow finding an explicit formula for a separable initial-value problem?

Mixed Separable Initial-Value Problem Workshop

What You'll Learn

A Reliable Method for Separable Initial-Value Problems

Separation, Integration, and the Initial Condition

Why the Separation Method Works

Worked Examples