Further Maths differential equations worksheet answers to download as PDF

These Further Maths differential equations worksheet answers give a full worked solution to all 20 questions of the differential equations worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \(I=e^{3x}\). \(\dfrac d{dx}(ye^{3x})=6e^{3x}\), so \(ye^{3x}=2e^{3x}+A\) and \(y=2+Ae^{-3x}\).
2. Divide by \(x\): \(\dfrac{dy}{dx}+\dfrac2xy=4\). \(I=e^{2\ln x}=x^2\). \(\dfrac d{dx}(x^2y)=4x^2\), so \(x^2y=\dfrac43x^3+C\) and \(y=\dfrac43x+\dfrac C{x^2}\).
3. \(I=e^{\int\tan x\,dx}=e^{-\ln\cos x}=\sec x\). \(\dfrac d{dx}(y\sec x)=\cos x\sec x=1\), so \(y\sec x=x+C\) and \(y=(x+C)\cos x\).
4. \(I=e^{2x}\). \(\dfrac d{dx}(ye^{2x})=e^{x}\), so \(ye^{2x}=e^x+C\) and \(y=e^{-x}+Ce^{-2x}\). When \(x=0\): \(1+C=3\), \(C=2\). So \(y=e^{-x}+2e^{-2x}\).
Answer to Question 2
Method: for \(ay''+by'+cy=0\), solve the auxiliary equation \(am^2+bm+c=0\). Two real roots give \(Ae^{m_1x}+Be^{m_2x}\); a repeated root gives \((A+Bx)e^{mx}\); complex roots \(p\pm qi\) give \(e^{px}(A\cos qx+B\sin qx)\).
1. The auxiliary equation factorises. \(m^2-5m+6=(m-2)(m-3)=0\), distinct real roots 2 and 3: \(y=Ae^{2x}+Be^{3x}\) Check: \(e^{2x}\) gives \(4-10+6=0\).
2. The auxiliary equation is a perfect square. \(m^2+6m+9=(m+3)^2=0\), repeated root \(-3\): \(y=(A+Bx)e^{-3x}\) The factor \(x\) is needed because \(Ae^{-3x}+Be^{-3x}\) would contain only one constant.
3. The discriminant is negative, so the roots are complex with \(p=-2\) and \(q=3\). \(m=\dfrac{-4\pm\sqrt{16-52}}2=-2\pm3i\): \(y=e^{-2x}(A\cos3x+B\sin3x)\) The solution oscillates with decreasing amplitude.
4. Here the roots are purely imaginary. \(m^2+16=0\), \(m=\pm4i\) (so \(p=0\)): \(y=A\cos4x+B\sin4x\) This is undamped oscillation, as in simple harmonic motion.
Answer to Question 3
1. Try \(y=ax+b\): \(-4a+3ax+3b=6x\), so \(a=2\) and \(3b=8\), \(b=\tfrac83\). The CF has roots 1 and 3. General solution \(y=Ae^{x}+Be^{3x}+2x+\tfrac83\).
2. The roots of \(m^2+m-6=0\) are 2 and \(-3\), so \(e^{2x}\) is in the CF. Try \(y=kxe^{2x}\): \(y'=k(1+2x)e^{2x}\), \(y''=k(4+4x)e^{2x}\). Substituting, \(k(4+4x)+k(1+2x)-6kx=5k\), so \(5k=10\), \(k=2\). General solution \(y=Ae^{2x}+Be^{-3x}+2xe^{2x}\).
3. Try \(y=a\sin2x+b\cos2x\). \(y''=-4y\), so \(y''+9y=5y\), giving \(5a=1\) and \(b=0\). The CF is \(A\cos3x+B\sin3x\). General solution \(y=A\cos3x+B\sin3x+\tfrac15\sin2x\).
Answer to Question 4
Method: for \(ay''+by'+cy=0\), solve the auxiliary equation \(am^2+bm+c=0\). Two real roots give \(Ae^{m_1x}+Be^{m_2x}\); a repeated root gives \((A+Bx)e^{mx}\); complex roots \(p\pm qi\) give \(e^{px}(A\cos qx+B\sin qx)\).
1. \(m^2-m-2=(m-2)(m+1)=0\), so \(m=2\) or \(m=-1\).
2. Two distinct real roots give a sum of two exponentials. \(y=Ae^{2x}+Be^{-x}\).
3. Differentiate the general solution before substituting \(x=0\) into \(y'\). \(y(0)=A+B=1\). \(y'=2Ae^{2x}-Be^{-x}\), so \(y'(0)=2A-B=5\). Adding: \(3A=6\), \(A=2\), \(B=-1\). \(y=2e^{2x}-e^{-x}\). Check: \(y(0)=2-1=1\) and \(y'(0)=4+1=5\).
4. Use \(e^{2\ln2}=4\) and \(e^{-\ln2}=\tfrac12\). \(y=2\cdot4-\tfrac12=\) \(7.5\) So \(y(\ln2)=7.5\).
Answer to Question 5
1. \(m^2+16=0\), so \(x=A\cos4t+B\sin4t\). \(x(0)=A=3\) and \(x'(0)=4B=0\). So \(x=3\cos4t\).
2. Amplitude 3 m; period \(\dfrac{2\pi}4=\) \(\dfrac\pi2\approx1.57\) s.
3. \(\dot x=-12\sin4t\), so the greatest speed is 12 m s\(^{-1}\) (when passing through the centre).
4. \(v^2=\omega^2(a^2-x^2)=16(9-4)=80\), so \(v=\) \(4\sqrt5\approx8.94\) m s\(^{-1}\).
Answer to Question 6
1. \(y'=-2e^{-2x}\), \(y''=4e^{-2x}\). Then \(4e^{-2x}-8e^{-2x}+4e^{-2x}=\) 0.
2. \(y'=(1-2x)e^{-2x}\), \(y''=(4x-4)e^{-2x}\). Then \((4x-4)+4(1-2x)+4x=0\), so it is a solution.
3. The auxiliary equation has the repeated root \(-2\), so \(y=(A+Bx)e^{-2x}\). A second-order equation needs two independent solutions to fit two initial conditions.
4. \(y(0)=A=1\). \(y'=(B-2A-2Bx)e^{-2x}\), so \(y'(0)=B-2=0\), \(B=2\). \(y=(1+2x)e^{-2x}\).
Answer to Question 7
Method: the substitution \(\theta=T-20\) turns the law of cooling into \(\dfrac{d\theta}{dt}=-k\theta\), whose solution is an exponential decay. Use the two data points to find the constant and \(k\).
1. Separate variables or use \(\theta=T-20\): \(\dfrac{d\theta}{dt}=-k\theta\), so \(\theta=\theta_0e^{-kt}\). With \(\theta_0=70\): \(T=20+70e^{-kt}\).
2. \(60=20+70e^{-5k}\) gives \(e^{-5k}=\tfrac47\), so \(5k=\ln\tfrac74\) and \(k=\tfrac15\ln\tfrac74=\) 0.112 (3 s.f.).
3. \(40=20+70e^{-kt}\), \(e^{-kt}=\tfrac27\), \(t=\dfrac{\ln3.5}{k}=\dfrac{1.2528}{0.11192}\approx\) 11.2 minutes.
4. \(e^{-kt}\to0\), so \(T\to20^\circ\)C, the room temperature.
Answer to Question 8
Method (integrating factor): write the equation as \(\dfrac{dy}{dx}+Py=Q\), multiply by \(I=e^{\int P\,dx}\); the left-hand side becomes \(\dfrac{d}{dx}(Iy)\). Integrate both sides, remember the constant, then divide by \(I\).
1. Move the \(C\) term to the left; \(P=0.2\), so \(I=e^{\int0.2\,dt}\). \(\dfrac{dC}{dt}+0.2C=4\). \(I=\) \(e^{0.2t}\).
2. Multiply through by \(I\) and integrate: \(\int4e^{0.2t}dt=20e^{0.2t}\). \(\dfrac d{dt}(Ce^{0.2t})=4e^{0.2t}\), so \(Ce^{0.2t}=20e^{0.2t}+K\). With \(C(0)=0\): \(K=-20\). So \(C=20(1-e^{-0.2t})\). This is the required result.
3. Look at the exponential term as \(t\) grows. \(e^{-0.2t}\to0\), so \(C\to\) 20 mg per litre. This is where the drip rate balances the rate at which the body removes the drug.
4. Set \(C=15\) and solve with logarithms. \(15=20(1-e^{-0.2t})\), \(e^{-0.2t}=\tfrac14\), \(t=5\ln4=\) 6.93 hours. So the target concentration is reached after about 6.93 hours.
Answer to Question 9
1. Rate in \(=0.1\times4=0.4\) kg per minute. Concentration in the tank is \(\dfrac Q{200}\), so rate out \(=4\times\dfrac Q{200}=\dfrac Q{50}\). Net rate: \(\dfrac{dQ}{dt}=0.4-\dfrac Q{50}\).
2. \(\dfrac{dQ}{dt}+\dfrac Q{50}=0.4\), \(I=e^{t/50}\), \(Qe^{t/50}=20e^{t/50}+K\). \(Q(0)=10\) gives \(K=-10\): \(Q=20-10e^{-t/50}\).
3. \(15=20-10e^{-t/50}\), \(e^{-t/50}=\tfrac12\), \(t=50\ln2=\) 34.7 minutes.
4. \(Q\to\) 20 kg, which is \(0.1\times200\): the tank ends up with the same concentration (0.1 kg per litre) as the incoming brine.
Answer to Question 10
1. \(RC=2000\times500\times10^{-6}=1\) second. So \(\dfrac{dV}{dt}+V=12\).
2. \(I=e^t\), \(Ve^t=12e^t+K\), \(V(0)=0\) gives \(K=-12\): \(V=12(1-e^{-t})\).
3. \(9=12(1-e^{-t})\), \(e^{-t}=\tfrac14\), \(t=\ln4\approx\) 1.39 s.
4. \(\dfrac{dV}{dt}=12e^{-t}\), which equals 12 V s\(^{-1}\) at \(t=0\): the voltage rises fastest at the start and slows as the capacitor fills.
Answer to Question 11
1. \(2m^2+8m+26=0\), so \(m^2+4m+13=0\), \(m=\dfrac{-4\pm\sqrt{16-52}}2=-2\pm3i\). The roots are complex, so the damping is light (underdamped).
2. \(x=e^{-2t}(A\cos3t+B\sin3t)\). \(x(0)=A=\tfrac12\). \(\dot x=e^{-2t}\left[(-2A+3B)\cos3t+(-2B-3A)\sin3t\right]\) and \(\dot x(0)=-2A+3B=0\), so \(B=\tfrac13\). \(x=e^{-2t}\left(\tfrac12\cos3t+\tfrac13\sin3t\right)\).
3. Period \(=\dfrac{2\pi}3\approx\) 2.09 s. Each period multiplies the amplitude by \(e^{-2\times2\pi/3}=e^{-4\pi/3}\approx\) 0.0152.
4. With \(c=20\): \(m^2+10m+13=0\), discriminant \(100-52=48>0\), two distinct negative real roots: the damping is heavy and the scooter returns to equilibrium without oscillating.
Answer to Question 12
1. Set \(\dfrac{dA}{dt}=0\): \(A=\dfrac{2000}{0.04}=\) £50 000.
2. \(\dfrac{dA}{dt}-0.04A=-2000\), \(I=e^{-0.04t}\), \(Ae^{-0.04t}=50000e^{-0.04t}+K\), so \(A=50000+Ke^{0.04t}\). \(A(0)=40000\) gives \(K=-10000\). \(A=50000-10000e^{0.04t}\).
3. \(A=0\) when \(e^{0.04t}=5\), \(t=25\ln5\approx\) 40.2 years.
4. £50 000, since then \(A'=0\) at all times. Starting below it, the balance falls ever faster.
Answer to Question 13
1. \(m^2+4=0\) gives \(m=\pm2i\): \(x_{CF}=A\cos2t+B\sin2t\).
2. \(x''=-9a\cos3t\), so \(-9a+4a=10\), \(a=-2\). General solution \(x=A\cos2t+B\sin2t-2\cos3t\).
3. The forcing has the same frequency as the CF, so \(a\cos2t+b\sin2t\) gives zero on the left side. For \(x=2t\sin2t\): \(\dot x=2\sin2t+4t\cos2t\) and \(\ddot x=8\cos2t-8t\sin2t\). Then \(\ddot x+4x=8\cos2t\). ✓.
4. The factor \(t\) means the amplitude grows without limit: resonance. If walkers match the natural frequency of the bridge, the swaying increases steadily, so soldiers break step to avoid that.
Answer to Question 14
Method (integrating factor): write the equation as \(\dfrac{dy}{dx}+Py=Q\), multiply by \(I=e^{\int P\,dx}\); the left-hand side becomes \(\dfrac{d}{dx}(Iy)\). Integrate both sides, remember the constant, then divide by \(I\).
1. At terminal speed the acceleration is zero. Terminal speed when \(\dfrac{dv}{dt}=0\): \(v=\dfrac{9.8}{0.2}=\) 49 m s\(^{-1}\). This is where air resistance balances the weight.
2. \(\dfrac{dv}{dt}+0.2v=9.8\), \(I=e^{0.2t}\), \(ve^{0.2t}=49e^{0.2t}+K\), \(K=-49\) (since \(v(0)=0\)). So \(v=49(1-e^{-0.2t})\). This is the required result.
3. Substitute \(t=10\). \(v(10)=49(1-e^{-2})=49\times0.8647=\) 42.4 m s\(^{-1}\). So after 10 s the skydiver is at about 87% of terminal speed.
4. Distance is the integral of speed. \(s=\int_0^{10}49(1-e^{-0.2t})\,dt=49\left[t+5e^{-0.2t}\right]_0^{10}=49(10+5e^{-2}-5)=\) 278 m (3 s.f.). So the skydiver falls about 278 m in the first 10 seconds.
Answer to Question 15
Method (integrating factor): write the equation as \(\dfrac{dy}{dx}+Py=Q\), multiply by \(I=e^{\int P\,dx}\); the left-hand side becomes \(\dfrac{d}{dx}(Iy)\). Integrate both sides, remember the constant, then divide by \(I\).
1. Divide through by \(x\) so that the coefficient of \(\dfrac{dy}{dx}\) is 1; then \(P=-\dfrac2x\). \(\dfrac{dy}{dx}-\dfrac2xy=x^3\). \(I=e^{-2\ln x}=\) \(x^{-2}\). Using \(e^{-2\ln x}=e^{\ln x^{-2}}=x^{-2}\).
2. Multiply by \(x^{-2}\); the left-hand side is the derivative of \(\dfrac y{x^2}\). \(\dfrac d{dx}\left(\dfrac y{x^2}\right)=x\), so \(\dfrac y{x^2}=\dfrac{x^2}2+C\): \(y=\dfrac{x^4}2+Cx^2\). Check: \(xy'-2y=x(2x^3+2Cx)-2\left(\tfrac{x^4}2+Cx^2\right)=x^4\).
3. Substitute \(x=1\), \(y=3\). \(3=\dfrac12+C\), \(C=\dfrac52\): \(y=\dfrac{x^4}2+\dfrac52x^2\).
4. \(y(2)=8+10=\) 18. \(y'=2x^3+5x>0\) for \(x>0\), so there is no stationary point. So the curve is always increasing for \(x>0\).
Answer to Question 16
1. \(m^2+2m+5=0\), \(m=-1\pm2i\): \(y_{CF}=e^{-x}(A\cos2x+B\sin2x)\).
2. \(y'=a\), \(y''=0\): \(2a+5ax+5b=5x+12\). So \(a=1\) and \(2+5b=12\), \(b=2\). \(y_{PI}=x+2\).
3. \(y=e^{-x}(A\cos2x+B\sin2x)+x+2\)
4. \(y(0)=A+2=3\), \(A=1\). \(y'=e^{-x}[(-A+2B)\cos2x+(-B-2A)\sin2x]+1\), and \(y'(0)=-A+2B+1=0\) gives \(B=0\). So \(y=e^{-x}\cos2x+x+2\). As \(x\) grows the oscillating term dies away and \(y\approx x+2\).
Answer to Question 17
1. Critical damping when the discriminant is zero: \(c^2-100=0\), so \(c=10\).
2. \(m^2+6m+25=0\), \(m=-3\pm4i\). The roots are complex, so the motion oscillates with decaying amplitude.
3. \(m^2+12m+25=0\), discriminant \(144-100=44>0\), \(m=-6\pm\sqrt{11}\). \(x=Ae^{(-6+\sqrt{11})t}+Be^{(-6-\sqrt{11})t}\).
4. \(x=(A+Bt)e^{-5t}\). \(x(0)=A=0\), \(\dot x=Be^{-5t}(1-5t)\), \(\dot x(0)=B=2\). \(x=2te^{-5t}\). \(\dot x=0\) at \(t=0.2\), so the greatest displacement is \(0.4e^{-1}\approx\) 0.147 m.
Answer to Question 18
1. \(I=e^x\). Multiplying: \(e^x\dfrac{dy}{dx}+ye^x=x\), which is \(\dfrac d{dx}(ye^x)=x\).
2. \(ye^x=\tfrac12x^2+C\). \(y(0)=1\) gives \(C=1\). \(y=\left(\tfrac12x^2+1\right)e^{-x}\).
3. Product rule: \(\dfrac{dy}{dx}=xe^{-x}-\left(\tfrac12x^2+1\right)e^{-x}=e^{-x}\left(x-\tfrac12x^2-1\right)=\) \(-\tfrac12(x-1)^2e^{-x}\).
4. The gradient is never positive and is zero only at \(x=1\). The curve decreases for all \(x\ge0\) and has a stationary point of inflection at \(x=1\), with \(y=1.5e^{-1}\approx0.552\). It tends to 0 as \(x\to\infty\).
Answer to Question 19
1. From the first equation \(2y=\dot x+x\). Differentiate: \(2\dot y=\ddot x+\dot x\). Substitute \(\dot y=-2x-y\): \(\ddot x+\dot x=-4x-2y=-4x-(\dot x+x)\), so \(\ddot x+2\dot x+5x=0\).
2. \(m=-1\pm2i\), \(x=e^{-t}(A\cos2t+B\sin2t)\). \(x(0)=1\) gives \(A=1\). \(\dot x(0)=-x(0)+2y(0)=-1\) and \(\dot x(0)=-A+2B\), so \(B=0\). \(x=e^{-t}\cos2t\).
3. \(\dot x=-e^{-t}\cos2t-2e^{-t}\sin2t\), so \(y=\tfrac12(\dot x+x)=\) \(-e^{-t}\sin2t\).
4. \(x^2+y^2=e^{-2t}\), so the point stays on a circle of radius \(e^{-t}\) that shrinks: it spirals clockwise into the origin (at \(t=\tfrac\pi4\) it is at \(-e^{-\pi/4}\) on the \(y\)-axis).
Answer to Question 20
1. \(\dfrac{dE}{dt}=\dot x\ddot x+\omega^2x\dot x=\dot x(\ddot x+\omega^2x)=\dot x(-2\lambda\dot x)=\) \(-2\lambda\dot x^2\).
2. If \(\lambda=0\) then \(\dfrac{dE}{dt}=0\), so \(E\) is constant. If \(\lambda>0\) then \(\dfrac{dE}{dt}\le0\) and the energy falls (friction removes it).
3. \(\dot x=-12\sin4t\), so \(E=72\sin^24t+\tfrac12\cdot16\cdot9\cos^24t=72\sin^24t+72\cos^24t=\) 72, constant.
4. Here \(\ddot x+4\dot x+13x=0\) (the scooter equation divided by 2), so \(2\lambda=4\). Evaluate \(\dot x(1)\) and \(E'(1)\) numerically: both sides equal \(-4\dot x(1)^2\approx-0.0068\). ✓

How to use these Further Maths differential equations worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the differential equations lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Further Maths differential equations worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Further Maths differential equations worksheet answers show every step, so nothing is left unexplained.