P4 6.4 - Solving separable differential equations | Edexcel A-Level International Mathematics

Recover a function from a differential equation by separating its variables and integrating. Distinguish general from particular solutions, keep solutions that division might lose, and use initial conditions and domain restrictions to interpret a model.

Separating variables

If a gradient is the product $xy^2$ , can you collect all the $y$ -dependence on one side and all the $x$ -dependence on the other? That rearrangement turns the equation into two integrals you already know.

A differential equation relates a function to its derivative. A solution is a function that satisfies the equation throughout an interval. Substituting a proposed solution and its derivative checks both sides at every input, not just at one point.

Separable differential equation

An equation of the form

\frac{dy}{dx}=f(x)g(y)

is separable. Where $g(y)\ne0$ , write

\frac1{g(y)}\,dy=f(x)\,dx

and integrate both sides. Values satisfying $g(y)=0$ must be checked separately as possible constant solutions before dividing.

The differential notation records a chain-rule rearrangement. After integrating, use one arbitrary constant: constants from the two sides combine into their difference.

Worked example 1: separate a product

Separate and integrate $dy/dx=xy$ .

Plan. Check $y=0$ , then divide by $y$ on any interval where it is nonzero.

The constant function $y=0$ gives $0=x\times0$ , so it is a solution. For nonzero $y$ ,

\frac1y\,dy=x\,dx \quad\Rightarrow\quad \int\frac1y\,dy=\int x\,dx \quad\Rightarrow\quad \ln|y|=\frac{x^2}{2}+C.

Answer. $\ln|y|=x^2/2+C$ for nonzero solutions, together with $y=0$ .

Check. Differentiate the implicit relation: $(1/y)(dy/dx)=x$ , giving the original $dy/dx=xy$ .

Change only the power of $y$ from $1$ to $2$ . Separation is identical, but the left integral uses a power instead of a logarithm.

Worked example 2: a power on the dependent variable

Separate and integrate $dy/dx=xy^2$ .

Plan. Retain $y=0$ , then collect $y^{-2}\,dy$ on the left.

y^{-2}\,dy=x\,dx \quad\Rightarrow\quad -\frac1y=\frac{x^2}{2}+C.

The power rule raises $-2$ to $-1$ and divides by $-1$ .

Answer. $-1/y=x^2/2+C$ , for nonzero $y$ , together with $y=0$ .

Check. Implicit differentiation gives $y^{-2}(dy/dx)=x$ , hence $dy/dx=xy^2$ . The constant zero solution also satisfies the equation directly.

The nonzero solutions can be made explicit later; first make sure the separation and integration are correct. This answers the opening question without losing the constant solution.

Watch out

$\int y^{-2}\,dy=-1/y$ , not $\ln|y^2|$ . Also, dividing by $y^2$ excludes $y=0$ , so check it in the original equation before dividing.

Change only the coefficient on the right.

General solutions

A general solution contains an arbitrary constant, so it describes a family of solution curves rather than one curve.

After logarithmic integration, $\ln|y|=H(x)+C$ gives $|y|=e^Ce^{H(x)}$ . Since $e^C>0$ , removing the absolute value allows either sign: write $y=Ae^{H(x)}$ , with nonzero real $A$ . A separately verified zero solution may let you include $A=0$ too. Do not replace $e^C$ by $C$ without explaining that you are renaming a constant.

Worked example 3: an explicit family

Find the general solution of $dy/dx=2xy$ .

Plan. Separate nonzero solutions, integrate, then include the checked constant solution.

\frac1y\,dy=2x\,dx \quad\Rightarrow\quad \ln|y|=x^2+C \quad\Rightarrow\quad y=Ae^{x^2}.

Both signs of $A$ represent nonzero solutions. The constant solution $y=0$ is obtained by allowing $A=0$ .

Answer. $y=Ae^{x^2}$ , for any real constant $A$ .

Check. $dy/dx=2xAe^{x^2}=2xy$ , including $A=0$ .

Change only the dependent-variable factor from $y$ to $y-3$ . The family is now shifted vertically.

Worked example 4: a shifted equilibrium

Find the general solution of $dy/dx=2x(y-3)$ .

Plan. Check $y=3$ , then separate using $y-3$ .

For nonconstant solutions,

\frac1{y-3}\,dy=2x\,dx \quad\Rightarrow\quad \ln|y-3|=x^2+C \quad\Rightarrow\quad y-3=Ae^{x^2}.

The checked constant solution $y=3$ is included by $A=0$ .

Answer. $y=3+Ae^{x^2}$ , for any real $A$ .

Check. Its derivative is $2xAe^{x^2}=2x(y-3)$ . Also, a nonzero $A$ keeps $y-3$ on one side of zero, consistent with the division.

Some general solutions are best left implicitly, for example $-1/y=x^2/2+C$ . If made explicit as $y=-1/(x^2/2+C)$ , exclude zeros of the denominator and retain $y=0$ separately. A formula is a solution only where the original equation and the function are defined.

Watch out

$\ln|y-3|=x^2+C$ does not give $y-3=e^{x^2}+C$ . Exponentiating a sum produces a product: $e^{x^2+C}=e^Ce^{x^2}$ .

Change the equilibrium value in the next equation.

Particular solutions

An initial condition selects one curve by fixing the arbitrary constant. Substitute the condition into the integrated equation before rearranging if that is simpler.

Domain restrictions remain after the constant is found. A denominator may introduce a new excluded point. When an initial point is given, use an interval containing that point that crosses neither an original exclusion nor a new one.

Worked example 5: a denominator limits the solution

Solve $dy/dx=y^2/(x+1)$ , given $y=1$ at $x=0$ , for $x>-1$ .

Plan. Separate and integrate, use the initial values to find the constant, then inspect the denominator.

\frac1{y^2}\,dy=\frac1{x+1}\,dx \quad\Rightarrow\quad -\frac1y=\ln(x+1)+C.

At $(0,1)$ , $-1=0+C$ , so

y=\frac1{1-\ln(x+1)}.

The denominator is zero at $x=e-1$ .

Answer. $y=1/[1-\ln(x+1)]$ , on the largest interval containing the initial point, $-1<x<e-1$ .

Check. At $x=0$ , $y=1$ . Differentiation gives $dy/dx=1/[(x+1)(1-\ln(x+1))^2]=y^2/(x+1)$ .

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P4 6.4 - Solving separable differential equations

Separating variables

General solutions

Particular solutions

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