Solving a separable differential equation | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Why separation produces two integrals

The complete solve order

1 Separatey with dy; x with dx

2 Integrateboth sides once

3 Use conditionfind the constant

4 Presentrequested form + check

Integrate before using the condition.

Do these four jobs in order; do not substitute the condition before integrating.

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

\frac{1}{g(y)}\,dy=f(x)\,dx

This notation is a compact form of a substitution based on the chain rule:

\int\frac{1}{g(y)}\frac{dy}{dx}\,dx=\int f(x)\,dx

\int\frac{1}{g(y)}\,dy=\int f(x)\,dx

It is therefore safe to use the familiar “y terms with dy, x terms with dx” layout, but $dy/dx$ is still a derivative, not an ordinary fraction that can always be split.

Use the four jobs in order

Separate: put every y-expression with $dy$ and every x-expression with $dx$ .
Integrate: use one integral on each side and write one constant.
Use the condition: substitute the given point to determine the constant.
Present: rearrange into the requested form and check the condition.

Worked example

Solve a basic separable equation completely

Solve $\dfrac{dy}{dx}=xy$ , given that $y=2$ when $x=0$ .

\frac1y\,dy=x\,dx

\ln|y|=\frac{x^2}{2}+C

The condition has $y>0$ , so this solution stays on the positive branch near the initial point.

\ln y=\frac{x^2}{2}+C

y=Ae^{x^2/2}

Substituting $x=0$ and $y=2$ gives $A=2$ .

Answer

y=2e^{x^2/2}

Common mistake

e^C

is another constant, but it is positive. If the sign is not known, write a non-zero constant such as A after removing the modulus; do not silently assume it is positive.

Verify the equation and the condition

Differentiate the final answer. It must reproduce the original differential equation.
Substitute the initial values. They must reproduce the given condition exactly.
State the connected interval when a function such as tangent has vertical asymptotes. A connected interval is one unbroken interval containing the initial x-value.

Your turnIndependent solution check [6 marks]

Solve $\dfrac{dy}{dx}=2x(1+y^2)$ , given that $y=0$ when $x=0$ . Give y in terms of x.

Show worked answer

\frac{1}{1+y^2}\,dy=2x\,dx

\tan^{-1}y=x^2+C

The condition gives $C=0$ .

Answer

y=\tan(x^2)

|x|<\sqrt{\frac{\pi}{2}}

Why separation produces two integrals

The complete solve order

1 Separatey with dy; x with dx

2 Integrateboth sides once

3 Use conditionfind the constant

4 Presentrequested form + check

Integrate before using the condition.

Do these four jobs in order; do not substitute the condition before integrating.

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

\frac{1}{g(y)}\,dy=f(x)\,dx

This notation is a compact form of a substitution based on the chain rule:

\int\frac{1}{g(y)}\frac{dy}{dx}\,dx=\int f(x)\,dx

\int\frac{1}{g(y)}\,dy=\int f(x)\,dx

It is therefore safe to use the familiar “y terms with dy, x terms with dx” layout, but $dy/dx$ is still a derivative, not an ordinary fraction that can always be split.

Use the four jobs in order

Separate: put every y-expression with $dy$ and every x-expression with $dx$ .
Integrate: use one integral on each side and write one constant.
Use the condition: substitute the given point to determine the constant.
Present: rearrange into the requested form and check the condition.

Worked example

Solve a basic separable equation completely

Solve $\dfrac{dy}{dx}=xy$ , given that $y=2$ when $x=0$ .

\frac1y\,dy=x\,dx

\ln|y|=\frac{x^2}{2}+C

The condition has $y>0$ , so this solution stays on the positive branch near the initial point.

\ln y=\frac{x^2}{2}+C

y=Ae^{x^2/2}

Substituting $x=0$ and $y=2$ gives $A=2$ .

Answer

y=2e^{x^2/2}

Common mistake

e^C

is another constant, but it is positive. If the sign is not known, write a non-zero constant such as A after removing the modulus; do not silently assume it is positive.

Verify the equation and the condition

Differentiate the final answer. It must reproduce the original differential equation.
Substitute the initial values. They must reproduce the given condition exactly.
State the connected interval when a function such as tangent has vertical asymptotes. A connected interval is one unbroken interval containing the initial x-value.

Your turnIndependent solution check [6 marks]

Solve $\dfrac{dy}{dx}=2x(1+y^2)$ , given that $y=0$ when $x=0$ . Give y in terms of x.

Show worked answer

\frac{1}{1+y^2}\,dy=2x\,dx

\tan^{-1}y=x^2+C

The condition gives $C=0$ .

Answer

y=\tan(x^2)

|x|<\sqrt{\frac{\pi}{2}}