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IB Mathematics: Applications and Interpretation HL · AHL 9: Differential Equations · AHL 9.2

Separation of variables: general and particular solutions Problem set B Worked solutions

Skills

  1. Find the general solution of each differential equation. Where the question gives the sign of \(y\), use it to remove the modulus.

    1. (a) \(\dfrac{dy}{dx}=-3y\), \(y>0\)

      • \(\int\dfrac1y\,dy=\int-3\,dx\)
      • \(\ln y=-3x+c\)
      • \(y=Ae^{-3x}\)
    2. (b) \(\dfrac{dy}{dx}=0.4(y-5)\), \(y>5\)

      • \(\int\dfrac{1}{y-5}\,dy=\int0.4\,dx\)
      • \(\ln(y-5)=0.4x+c\)
      • \(y=5+Ae^{0.4x}\)
    3. (c) \(\dfrac{dy}{dx}=2(9-y)\), \(y<9\)

      • \(\int\dfrac{1}{9-y}\,dy=\int2\,dx\)
      • \(-\ln(9-y)=2x+c\)
      • \(9-y=Ae^{-2x},\ \ y=9-Ae^{-2x}\)
  2. Find the general solution of each differential equation. Give \(y\) in terms of \(x\).

    1. (a) \(\dfrac{dy}{dx}=6xy^2\)

      • \(\int\dfrac{1}{y^2}\,dy=\int6x\,dx\)
      • \(-\dfrac1y=3x^2+c\)
      • \(y=-\dfrac{1}{3x^2+c}\)
    2. (b) \(\dfrac{dy}{dx}=\dfrac{5y}{x}\), \(x>0,\ y>0\)

      • \(\int\dfrac1y\,dy=\int\dfrac5x\,dx\)
      • \(\ln y=5\ln x+c\)
      • \(y=Ax^5\)
    3. (c) \(y\dfrac{dy}{dx}=7x^2\), \(y>0\)

      • \(\int y\,dy=\int7x^2\,dx\)
      • \(\dfrac{y^2}{2}=\dfrac{7x^3}{3}+c\)
      • \(y^2=\dfrac{14x^3}{3}+A,\ \ y=\sqrt{\dfrac{14x^3}{3}+A}\)
  3. Find the general solution of each differential equation. Give \(y\) in terms of \(x\).

    1. (a) \(\dfrac{dy}{dx}=e^{-y}\cos x\)

      • \(\int e^{y}\,dy=\int\cos x\,dx\)
      • \(e^{y}=\sin x+c\)
      • \(y=\ln(\sin x+A)\)
    2. (b) \(\dfrac{dy}{dx}=ye^{-x}\), \(y>0\)

      • \(\int\dfrac1y\,dy=\int e^{-x}\,dx\)
      • \(\ln y=-e^{-x}+c\)
      • \(y=Ae^{-e^{-x}}\)
  4. Find the particular solution of each differential equation.

    1. (a) \(\dfrac{dy}{dx}=5xy\), \(y=3\) when \(x=0\); give \(y\) in terms of \(x\).

      • \(\ln y=\dfrac{5x^2}{2}+c\ \Rightarrow\ y=Ae^{5x^2/2}\)
      • \(3=Ae^{0}\ \Rightarrow\ A=3\)
      • \(y=3e^{5x^2/2}\)
    2. (b) \(y^2\dfrac{dy}{dx}=6x\), \(y=2\) when \(x=1\); give \(y\) in terms of \(x\).

      • \(\dfrac{y^3}{3}=3x^2+c\)
      • \(\dfrac{8}{3}=3+c\ \Rightarrow\ c=-\dfrac13\)
      • \(y^3=9x^2-1,\ \ y=\sqrt[3]{9x^2-1}\)

Stretch

  1. Given that \(\dfrac{dy}{dx}=2x(y+1)\) and that \(y=3\) when \(x=0\), show that \(y=4e^{x^2}-1\).

    • \(\int\dfrac{1}{y+1}\,dy=\int2x\,dx\)
    • \(\ln|y+1|=x^2+c\)
    • \(y+1=Ae^{x^2}\)
    • \(x=0,\ y=3:\ 4=A\)
    • \(y=4e^{x^2}-1\)
  2. Given that \(\dfrac{dy}{dx}=e^{x}\cos^2y\), use separation of variables to show that \(\tan y=e^{x}+c\) for some constant \(c\).

    • \(\int\dfrac{1}{\cos^2y}\,dy=\int e^{x}\,dx\)
    • \(\int\sec^2y\,dy=\int e^{x}\,dx\)
    • \(\tan y=e^{x}+c\)
  3. Given that \(\dfrac{dy}{dx}=\dfrac{y^2}{x^2}\), \(x>0\) and that \(y=0.5\) when \(x=1\), find \(y\) in terms of \(x\).

    • \(\int\dfrac{1}{y^2}\,dy=\int\dfrac{1}{x^2}\,dx\)
    • \(-\dfrac1y=-\dfrac1x+c\)
    • \(x=1,\ y=0.5:\ -2=-1+c\Rightarrow c=-1\)
    • \(\dfrac1y=\dfrac1x+1=\dfrac{1+x}{x}\)
    • \(y=\dfrac{x}{x+1}\)

jmaths.xyz · AI HL · AHL 9.2 Separation of variables: general and particular solutions