Further Maths First-order differential equations Practice Questions

Free Further Maths First-order differential equations practice questions with full step-by-step worked solutions. Covers integrating-factor, linear-first-order, separable, first-order. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Find an integrating factor for the differential equation dydx+3y=e2x\frac{dy}{dx}+3y=\mathrm{e}^{2x}.
Show worked solution

Worked solution

  1. Identify P(x)P\left(x\right) and Q(x)Q\left(x\right)

    P(x)=3,Q(x)=e2xP\left(x\right)=3,\qquad Q\left(x\right)=\mathrm{e}^{2x}

    PP is the coefficient of yy once the equation is in standard form.

  2. Form the integrating factor

    μ=e∫3 dx=e3x\mu=\mathrm{e}^{\int 3\,dx}=\mathrm{e}^{3x}

    This is the multiplier that makes the left-hand side exact.

  3. State the integrating factor

    μ=e3x\mu=\mathrm{e}^{3x}

    Multiplying the equation by this factor would make the left-hand side an exact derivative.

Answer
μ=e3x\mu=\mathrm{e}^{3x}
Question 2
2 markseasy
Which of the following is equal to x4dydx+4x3yx^{4}\frac{dy}{dx}+4x^{3}y?
Show worked solution

Worked solution

  1. Apply the product rule to the proposed derivative

    ddx(x4y)=x4dydx+4x3y\frac{d}{dx}\left(x^{4}y\right)=x^{4}\frac{dy}{dx}+4x^{3}y

    The product rule differentiates x4x^{4} and yy in turn.

  2. Compare with the given expression

    x4dydx+4x3y=x4dydx+4x3yx^{4}\frac{dy}{dx}+4x^{3}y=x^{4}\frac{dy}{dx}+4x^{3}y

    The two expressions match term for term.

  3. Reject option B

    ddx(4x3y)=4x3dydx+12x2y\frac{d}{dx}\left(4x^{3}y\right)=4x^{3}\frac{dy}{dx}+12x^{2}y

    This does not reproduce the given expression, so the option is wrong.

  4. Select the exact derivative

    ddx(x4y)\frac{d}{dx}\left(x^{4}y\right)

    The given expression is precisely the derivative of x4yx^{4}y.

Answer
ddx(x4y)\frac{d}{dx}\left(x^{4}y\right)
Question 3
4 marksintermediate
Which of the following is an integrating factor for the differential equation dydx+2xyx2+1=x\frac{dy}{dx}+\frac{2xy}{x^{2}+1}=x?
Show worked solution

Worked solution

  1. Identify P(x)P\left(x\right) and Q(x)Q\left(x\right)

    P(x)=2xx2+1,Q(x)=xP\left(x\right)=\frac{2x}{x^{2}+1},\qquad Q\left(x\right)=x

    PP is the coefficient of yy once the equation is in standard form.

  2. Integrate P(x)P\left(x\right)

    ∫2xx2+1 dx=ln⁡(x2+1)\int \frac{2x}{x^{2}+1}\,dx=\ln{\left(x^{2}+1\right)}

    No constant of integration is needed at this stage.

  3. Form the integrating factor

    μ=e∫2xx2+1 dx=eln⁡(x2+1)=x2+1\mu=\mathrm{e}^{\int \frac{2x}{x^{2}+1}\,dx}=\mathrm{e}^{\ln{\left(x^{2}+1\right)}}=x^{2}+1

    This is the multiplier that makes the left-hand side exact.

  4. Test the defining property

    dμdx=2x,P(x)μ=2x\frac{d\mu}{dx}=2x,\qquad P\left(x\right)\mu=2x

    The two agree, so this function really is an integrating factor.

  5. Reject option B

    ddx(ln⁡(x2+1))=2xx2+1≠2xln⁡(x2+1)x2+1\frac{d}{dx}\left(\ln{\left(x^{2}+1\right)}\right)=\frac{2x}{x^{2}+1}\neq \frac{2x\ln{\left(x^{2}+1\right)}}{x^{2}+1}

    This function does not satisfy dμdx=Pμ\frac{d\mu}{dx}=P\mu, so it is not an integrating factor.

  6. Reject option C

    ddx(e2x)=2e2x≠2xe2xx2+1\frac{d}{dx}\left(\mathrm{e}^{2x}\right)=2\mathrm{e}^{2x}\neq \frac{2x\mathrm{e}^{2x}}{x^{2}+1}

    This function does not satisfy dμdx=Pμ\frac{d\mu}{dx}=P\mu, so it is not an integrating factor.

  7. Select the integrating factor

    μ=x2+1\mu=x^{2}+1

    This is the only option satisfying dμdx=P(x)μ\frac{d\mu}{dx}=P\left(x\right)\mu.

Answer
x2+1x^{2}+1
Question 4
6 markshard
A tank holds 2525 litres of water and brine enters at 11 litre per minute carrying 88 grams of salt per litre, the well-mixed solution leaving at the same rate. The mass of salt in the tank is SS grams after tt minutes. The model gives the differential equation dSdt+S25=8\frac{dS}{dt}+\frac{S}{25}=8. Given that S=0S=0 when t=0t=0, which of the following best describes the behaviour of SS as t→∞t\to\infty?
Show worked solution

Worked solution

  1. Identify P(t)P\left(t\right) and Q(t)Q\left(t\right)

    P(t)=125,Q(t)=8P\left(t\right)=\frac{1}{25},\qquad Q\left(t\right)=8

    PP is the coefficient of SS once the equation is in standard form.

  2. Integrate P(t)P\left(t\right)

    ∫125 dt=t25\int \frac{1}{25}\,dt=\frac{t}{25}

    No constant of integration is needed at this stage.

  3. Form the integrating factor

    μ=e∫125 dt=et25\mu=\mathrm{e}^{\int \frac{1}{25}\,dt}=\mathrm{e}^{\frac{t}{25}}

    This is the multiplier that makes the left-hand side exact.

  4. Recognise the left-hand side as an exact derivative

    ddt(et25S)=8et25\frac{d}{dt}\left(\mathrm{e}^{\frac{t}{25}}S\right)=8\mathrm{e}^{\frac{t}{25}}

    By the product rule the left-hand side is exactly the derivative of et25S\mathrm{e}^{\frac{t}{25}}S, which is the whole point of the integrating factor.

  5. Carry out the integration

    et25S=∫8et25 dt=200et25+C\mathrm{e}^{\frac{t}{25}}S=\int 8\mathrm{e}^{\frac{t}{25}}\,dt=200\mathrm{e}^{\frac{t}{25}}+C

    The constant of integration is introduced here, and here only.

  6. Divide through by the integrating factor

    S=200et25+Cet25S=\frac{200\mathrm{e}^{\frac{t}{25}}+C}{\mathrm{e}^{\frac{t}{25}}}

    This makes SS the subject and gives the general solution.

  7. Apply the boundary condition

    S=0 when t=0 ⇒ 0=C+200S=0\ \text{when}\ t=0\ \Rightarrow\ 0=C+200

    Substituting the given values turns the general solution into an equation for CC.

  8. Solve for the arbitrary constant

    C=−200C=-200

    This single value of CC selects the one curve through the given point.

  9. State the particular solution

    S=200−200e−t25S=200-200\mathrm{e}^{-\frac{t}{25}}

    The behaviour of the model is read directly from this expression.

  10. Take the limit of the transient term

    −200e−t25→0 as t→∞-200\mathrm{e}^{-\frac{t}{25}}\to 0\ \text{as}\ t\to\infty

    A negative exponential tends to zero, so only the steady part survives.

  11. State the limiting behaviour

    S→200 as t→∞S\to 200\ \text{as}\ t\to\infty

    The model therefore settles down to this limiting value.

Answer
S→200S\to 200
Question 5
9 markschallenging
An RL circuit driven by an alternating supply carries a current II amperes after tt seconds. Which of the following is the solution of the differential equation dIdt+3I=6cos⁡(t)\frac{dI}{dt}+3I=6\cos{\left(t\right)} for which I=0I=0 when t=0t=0?
Show worked solution

Worked solution

  1. Identify P(t)P\left(t\right) and Q(t)Q\left(t\right)

    P(t)=3,Q(t)=6cos⁡(t)P\left(t\right)=3,\qquad Q\left(t\right)=6\cos{\left(t\right)}

    PP is the coefficient of II once the equation is in standard form.

  2. Integrate P(t)P\left(t\right)

    ∫3 dt=3t\int 3\,dt=3t

    No constant of integration is needed at this stage.

  3. Form the integrating factor

    μ=e∫3 dt=e3t\mu=\mathrm{e}^{\int 3\,dt}=\mathrm{e}^{3t}

    This is the multiplier that makes the left-hand side exact.

  4. Multiply the equation through by the integrating factor

    e3tdIdt+3e3tI=6e3tcos⁡(t)\mathrm{e}^{3t}\frac{dI}{dt}+3\mathrm{e}^{3t}I=6\mathrm{e}^{3t}\cos{\left(t\right)}

    Every term of the equation is multiplied, so the equation is unchanged.

  5. Recognise the left-hand side as an exact derivative

    ddt(e3tI)=6e3tcos⁡(t)\frac{d}{dt}\left(\mathrm{e}^{3t}I\right)=6\mathrm{e}^{3t}\cos{\left(t\right)}

    By the product rule the left-hand side is exactly the derivative of e3tI\mathrm{e}^{3t}I, which is the whole point of the integrating factor.

  6. Integrate both sides with respect to tt

    e3tI=∫6e3tcos⁡(t) dt\mathrm{e}^{3t}I=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt

    Integrating an exact derivative simply undoes it.

  7. Integrate by parts twice

    J=∫6e3tcos⁡(t) dt⇒J=3(sin⁡(t)+3cos⁡(t))e3t5J=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt\quad\Rightarrow\quad J=\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}

    Applying parts twice reproduces JJ on the right-hand side, and the resulting equation is then solved for JJ.

  8. Carry out the integration

    e3tI=∫6e3tcos⁡(t) dt=3(sin⁡(t)+3cos⁡(t))e3t5+C\mathrm{e}^{3t}I=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt=\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}+C

    The constant of integration is introduced here, and here only.

  9. Check the integration by differentiating

    ddt(3(sin⁡(t)+3cos⁡(t))e3t5)=6e3tcos⁡(t)\frac{d}{dt}\left(\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}\right)=6\mathrm{e}^{3t}\cos{\left(t\right)}

    Differentiating the answer returns the integrand, so the integration is correct.

  10. Divide through by the integrating factor

    I=3(sin⁡(t)+3cos⁡(t))e3t5+Ce3tI=\frac{\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}+C}{\mathrm{e}^{3t}}

    This makes II the subject and gives the general solution.

  11. Apply the boundary condition

    I=0 when t=0 ⇒ 0=C+95I=0\ \text{when}\ t=0\ \Rightarrow\ 0=C+\frac{9}{5}

    Substituting the given values turns the general solution into an equation for CC.

  12. Solve for the arbitrary constant

    C=−95C=-\frac{9}{5}

    This single value of CC selects the one curve through the given point.

  13. Reject option B

    ddt(2sin⁡(t)5+6cos⁡(t)5−6e−3t5) gives a residual of −2cos⁡(t)≠0\frac{d}{dt}\left(\frac{2\sin{\left(t\right)}}{5}+\frac{6\cos{\left(t\right)}}{5}-\frac{6\mathrm{e}^{-3t}}{5}\right)\ \text{gives a residual of}\ -2\cos{\left(t\right)}\neq 0

    Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.

  14. Reject option C

    ddt(9sin⁡(t)5+3cos⁡(t)5−3e−3t5) gives a residual of 24sin⁡(t)5−12cos⁡(t)5≠0\frac{d}{dt}\left(\frac{9\sin{\left(t\right)}}{5}+\frac{3\cos{\left(t\right)}}{5}-\frac{3\mathrm{e}^{-3t}}{5}\right)\ \text{gives a residual of}\ \frac{24\sin{\left(t\right)}}{5}-\frac{12\cos{\left(t\right)}}{5}\neq 0

    Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.

  15. Reject option D

    ddt(2sin⁡(t)−2e−3t) gives a residual of 6sin⁡(t)−4cos⁡(t)≠0\frac{d}{dt}\left(2\sin{\left(t\right)}-2\mathrm{e}^{-3t}\right)\ \text{gives a residual of}\ 6\sin{\left(t\right)}-4\cos{\left(t\right)}\neq 0

    Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.

  16. Select the correct solution

    I=3sin⁡(t)5+9cos⁡(t)5−9e−3t5I=\frac{3\sin{\left(t\right)}}{5}+\frac{9\cos{\left(t\right)}}{5}-\frac{9\mathrm{e}^{-3t}}{5}

    This is the only option with a zero residual in the differential equation.

Answer
I=3sin⁡(t)5+9cos⁡(t)5−9e−3t5I=\frac{3\sin{\left(t\right)}}{5}+\frac{9\cos{\left(t\right)}}{5}-\frac{9\mathrm{e}^{-3t}}{5}

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