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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 A Worked solutions
Skills
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Find the general solution of each differential equation. Where the question gives the sign of \(y\), use it to remove the modulus.
(a) \(\dfrac{dy}{dx}=5y\), \(y>0\)
- \(\int\dfrac1y\,dy=\int5\,dx\)
- \(\ln y=5x+c\)
- \(y=Ae^{5x}\)
(b) \(\dfrac{dy}{dx}=0.2(y-6)\), \(y>6\)
- \(\int\dfrac{1}{y-6}\,dy=\int0.2\,dx\)
- \(\ln(y-6)=0.2x+c\)
- \(y=6+Ae^{0.2x}\)
(c) \(\dfrac{dy}{dx}=4-y\), \(y<4\)
- \(\int\dfrac{1}{4-y}\,dy=\int1\,dx\)
- \(-\ln(4-y)=x+c\)
- \(4-y=Ae^{-x},\ \ y=4-Ae^{-x}\)
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Find the general solution of each differential equation. Give \(y\) in terms of \(x\).
(a) \(\dfrac{dy}{dx}=4xy^2\)
- \(\int\dfrac{1}{y^2}\,dy=\int4x\,dx\)
- \(-\dfrac1y=2x^2+c\)
- \(y=-\dfrac{1}{2x^2+c}\)
(b) \(\dfrac{dy}{dx}=\dfrac{3y}{x}\), \(x>0,\ y>0\)
- \(\int\dfrac1y\,dy=\int\dfrac3x\,dx\)
- \(\ln y=3\ln x+c\)
- \(y=Ax^3\)
(c) \(y\dfrac{dy}{dx}=5x^2\), \(y>0\)
- \(\int y\,dy=\int5x^2\,dx\)
- \(\dfrac{y^2}{2}=\dfrac{5x^3}{3}+c\)
- \(y^2=\dfrac{10x^3}{3}+A,\ \ y=\sqrt{\dfrac{10x^3}{3}+A}\)
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Find the general solution of each differential equation. Give \(y\) in terms of \(x\).
(a) \(\dfrac{dy}{dx}=e^{-y}\sin x\)
- \(\int e^{y}\,dy=\int\sin x\,dx\)
- \(e^{y}=-\cos x+c\)
- \(y=\ln(A-\cos x)\)
(b) \(\dfrac{dy}{dx}=ye^{2x}\), \(y>0\)
- \(\int\dfrac1y\,dy=\int e^{2x}\,dx\)
- \(\ln y=\dfrac{e^{2x}}{2}+c\)
- \(y=Ae^{e^{2x}/2}\)
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Find the particular solution of each differential equation.
(a) \(\dfrac{dy}{dx}=3xy\), \(y=2\) when \(x=0\); give \(y\) in terms of \(x\).
- \(\ln y=\dfrac{3x^2}{2}+c\ \Rightarrow\ y=Ae^{3x^2/2}\)
- \(2=Ae^{0}\ \Rightarrow\ A=2\)
- \(y=2e^{3x^2/2}\)
(b) \(y^2\dfrac{dy}{dx}=4x\), \(y=3\) when \(x=1\); give \(y\) in terms of \(x\).
- \(\dfrac{y^3}{3}=2x^2+c\)
- \(\dfrac{27}{3}=2+c\ \Rightarrow\ c=7\)
- \(y^3=6x^2+21,\ \ y=\sqrt[3]{6x^2+21}\)
Stretch
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Given that \(\dfrac{dy}{dx}=x(y-1)\) and that \(y=4\) when \(x=0\), show that \(y=1+3e^{x^2/2}\).
- \(\int\dfrac{1}{y-1}\,dy=\int x\,dx\)
- \(\ln|y-1|=\dfrac{x^2}{2}+c\)
- \(y-1=Ae^{x^2/2}\)
- \(x=0,\ y=4:\ 3=A\)
- \(y=1+3e^{x^2/2}\)
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Given that \(\dfrac{dy}{dx}=3x^2\cos^2y\), use separation of variables to show that \(\tan y=x^3+c\) for some constant \(c\).
- \(\int\dfrac{1}{\cos^2y}\,dy=\int3x^2\,dx\)
- \(\int\sec^2y\,dy=\int3x^2\,dx\)
- \(\tan y=x^3+c\)
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Given that \(\dfrac{dy}{dx}=\dfrac{y^2}{x^2}\), \(x>0\) and that \(y=1\) when \(x=2\), find \(y\) in terms of \(x\).
- \(\int\dfrac{1}{y^2}\,dy=\int\dfrac{1}{x^2}\,dx\)
- \(-\dfrac1y=-\dfrac1x+c\)
- \(x=2,\ y=1:\ -1=-\dfrac12+c\Rightarrow c=-\dfrac12\)
- \(\dfrac1y=\dfrac1x+\dfrac12=\dfrac{2+x}{2x}\)
- \(y=\dfrac{2x}{x+2}\)
jmaths.xyz · AI HL · AHL 9.2 Separation of variables: general and particular solutions