Menu
Sign up Sign in

AI HL · AHL 9

Differential Equations

0 of 66 skills done

0
AHL 9.2

Separation of variables: general and particular solutions

AHL 9.2

Separation of variables: general and particular solutions

Target: I can separate the variables in dy/dx = f(x)g(y) and integrate both sides, keeping one constant. (+4 more)

By the end of this section

Key points

The method

  • If \(\dfrac{dy}{dx}=f(x)g(y)\), put all the \(y\) on one side and all the \(x\) on the other: \(\displaystyle\int\dfrac{1}{g(y)}\,dy=\int f(x)\,dx\).
  • Put a constant of integration on one side only. Two constants are really one.
  • Treating \(\dfrac{dy}{dx}\) as a fraction is a good way to remember the method. What is really happening is \(\displaystyle\int\dfrac{1}{g(y)}\dfrac{dy}{dx}\,dx=\int f(x)\,dx\).

Getting y on its own

  • \(\displaystyle\int\dfrac{1}{y+a}\,dy=\ln|y+a|\). The question gives the sign of \(y+a\) (for example \(y>0\)), so you can drop the modulus.
  • Raise \(e\) to both sides. Write \(e^{c}\) as a new constant \(A\), a multiplier: \(y=Ae^{x^2/2}\).
  • Common error: writing \(y=e^{x^2/2}+c\). The constant multiplies, it is not added.
  • For \(\dfrac{dy}{dx}=k(y+a)\) the answer is \(y=Ae^{kx}-a\).

Particular solutions

  • Find the general solution first. Then substitute the pair of values \((x,y)\) the question gives to find the constant.
  • Give the answer in the form asked: \(y=\) a function of \(x\), or left as \(f(y)=g(x)\) when the question says you may.
  • When you take a root to get \(y\), the initial condition tells you whether to keep the positive or the negative root.

Show that

  • Start from the differential equation and separate the variables. Integrate, use the condition to find the constant, and finish on the form you were given.
  • Write every line. The marks are for the working that reaches the stated answer.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Equation of the form dy/dx = k(y + a)

Find the general solution of \(\dfrac{dy}{dx}=0.5(y+3)\), given that \(y>-3\). Write it as \(y=Ae^{0.5x}+a\) and give \(a\).

Working
  1. Separate

    \(\int\dfrac{1}{y+3}\,dy=\int0.5\,dx\)

    All the y on the left, all the x on the right.

  2. Integrate

    \(\ln|y+3|=0.5x+c\)

    One constant, on one side.

  3. Remove the modulus

    \(\ln(y+3)=0.5x+c\)

    y > −3, so y + 3 is positive.

  4. Exponentiate

    \(y+3=e^{c}e^{0.5x}=Ae^{0.5x}\)

    A = e^c is the new constant.

  5. Solve for y

    \(y=Ae^{0.5x}-3\)

    So a = −3.

Video walkthrough coming
Powers of x and y

Find the general solution of \(\dfrac{dy}{dx}=\dfrac{4x^3}{y^2}\) in the form \(y^3=px^4+A\). Give \(p\).

Working
  1. Separate

    \(\int y^2\,dy=\int4x^3\,dx\)

    Multiply both sides by y² and by dx.

  2. Integrate

    \(\dfrac{y^3}{3}=x^4+c\)

    One constant.

  3. Multiply by 3

    \(y^3=3x^4+3c=3x^4+A\)

    Every term is multiplied by 3, including the constant. Write 3c as A.

  4. Give y

    \(y=\sqrt[3]{3x^4+A}\)

    A cube root has one real value, so there is no ± here.

Video walkthrough coming
A trigonometric factor

Find the general solution of \(\dfrac{dy}{dx}=y^2\sin x\) in the form \(y=\dfrac{1}{\cos x+A}\). Then find \(A\) when \(y=0.25\) at \(x=0\).

Working
  1. Separate

    \(\int\dfrac{1}{y^2}\,dy=\int\sin x\,dx\)

    Divide by y².

  2. Integrate

    \(-\dfrac1y=-\cos x+c\)

    The integral of y⁻² is −1/y.

  3. Rearrange

    \(\dfrac1y=\cos x-c=\cos x+A\)

    Multiply by −1 and write −c as A.

  4. Invert

    \(y=\dfrac{1}{\cos x+A}\)

    This is the general solution.

  5. Use the condition

    \(4=\cos0+A\Rightarrow A=3\)

    1/0.25 = 4.

Video walkthrough coming
A particular solution

Given that \(\dfrac{dy}{dx}=\dfrac{x^2}{y}\) and that \(y=4\) when \(x=3\), find \(y\) in terms of \(x\). The general solution is \(\dfrac{y^2}{2}=\dfrac{x^3}{3}+c\). Find \(c\).

Working
  1. Separate and integrate

    \(\int y\,dy=\int x^2\,dx\ \Rightarrow\ \dfrac{y^2}{2}=\dfrac{x^3}{3}+c\)

    Given in the question.

  2. Substitute x = 3, y = 4

    \(\dfrac{16}{2}=\dfrac{27}{3}+c\ \Rightarrow\ 8=9+c\)

    Square the 4 first.

  3. Constant

    \(c=-1\)

  4. Give y

    \(y^2=\dfrac{2x^3}{3}-2,\quad y=\sqrt{\dfrac{2x^3}{3}-2}\)

    y = 4 is positive, so take the positive root.

Video walkthrough coming
Show that

Given that \(\dfrac{dy}{dx}=3x^2(y+2)\) and that \(y=1\) when \(x=0\), show that \(y=3e^{x^3}-2\). Give the value of \(A\) in \(y=Ae^{x^3}-2\).

Working
  1. Separate

    \(\int\dfrac{1}{y+2}\,dy=\int3x^2\,dx\)

  2. Integrate

    \(\ln|y+2|=x^3+c\)

  3. Exponentiate

    \(y+2=Ae^{x^3}\)

    A = e^c. Near the start y + 2 = 3 > 0, so the modulus can go.

  4. Use y = 1 at x = 0

    \(1+2=Ae^{0}\Rightarrow A=3\)

    e⁰ = 1.

  5. Finish

    \(y=3e^{x^3}-2\)

    This is the result to be shown.

Video walkthrough coming