A metal block at °C cools in a room whose temperature is constant; t is measured in minutes. The model gives the differential equation (d)/(dt)+()/(15)=(4)/(3). Given that =75 when t=0, which of the following best describes the behaviour of as t?

Identify P(t) and Q(t): P(t)=(1)/(15), Q(t)=(4)/(3) Integrate P(t): (1)/(15) dt=(t)/(15) Form the integrating factor: =e^ (1)/(15) dt=e^(t)/(15) Multiply the equation through by the integrating factor: e^(t)/(15)(d)/(dt)+e^(t)/(15)15=4e^(t)/(15)3 Recognise the left-hand side as an exact derivative: (d)/(dt)(e^(t)/(15))=4e^(t)/(15)3 Integrate both sides with respect to t: e^(t)/(15)= 4e^(t)/(15)3 dt Carry out the integration: e^(t)/(15)= 4e^(t)/(15)3 dt=20e^(t)/(15)+C Check the integration by differentiating: (d)/(dt)(20e^(t)/(15))=4e^(t)/(15)3 Divide through by the integrating factor: =20e^(t)/(15)+Ce^(t)/(15) Apply the boundary condition: =75 when t=0 75=C+20 Solve for the arbitrary constant: C=55 State the particular solution: =20+55e^-(t)/(15) Split the solution into a steady part and a transient part: _steady=20, _transient=55e^-(t)/(15) Take the limit of the transient term: 55e^-(t)/(15) 0 as t Differentiate the solution as a check: (d)/(dt)=-11e^-(t)/(15)3 Substitute the solution back into the equation: LHS=(4)/(3),=(4)/(3) State the limiting behaviour: 20 as t Answer: 20

Hard Further Maths First-order differential equations Questions

Challenging, exam-style Further Maths First-order differential equations questions with worked solutions. Stretch yourself on the hardest linear-first-order, integrating-factor, integration-by-parts, exact-equation problems.

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Further Maths34 questionsStep-by-step solutions

Question 1

9 markschallenging

An RL circuit driven by an alternating supply carries a current

I

amperes after

t

seconds. Which of the following is the solution of the differential equation

\frac{dI}{dt}+3I=6\cos{\left(t\right)}

for which

I=0

when

t=0

Show worked solution

Worked solution

Identify $P\left(t\right)$ and $Q\left(t\right)$
$P\left(t\right)=3,\qquad Q\left(t\right)=6\cos{\left(t\right)}$
$P$ is the coefficient of $I$ once the equation is in standard form.
Integrate $P\left(t\right)$
$\int 3\,dt=3t$
No constant of integration is needed at this stage.
Form the integrating factor
$\mu=\mathrm{e}^{\int 3\,dt}=\mathrm{e}^{3t}$
This is the multiplier that makes the left-hand side exact.
Multiply the equation through by the integrating factor
$\mathrm{e}^{3t}\frac{dI}{dt}+3\mathrm{e}^{3t}I=6\mathrm{e}^{3t}\cos{\left(t\right)}$
Every term of the equation is multiplied, so the equation is unchanged.
Recognise the left-hand side as an exact derivative
$\frac{d}{dt}\left(\mathrm{e}^{3t}I\right)=6\mathrm{e}^{3t}\cos{\left(t\right)}$
By the product rule the left-hand side is exactly the derivative of $\mathrm{e}^{3t}I$ , which is the whole point of the integrating factor.
Integrate both sides with respect to $t$
$\mathrm{e}^{3t}I=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt$
Integrating an exact derivative simply undoes it.
Integrate by parts twice
$J=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt\quad\Rightarrow\quad J=\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}$
Applying parts twice reproduces $J$ on the right-hand side, and the resulting equation is then solved for $J$ .
Carry out the integration
$\mathrm{e}^{3t}I=\int 6\mathrm{e}^{3t}\cos{\left(t\right)}\,dt=\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}+C$
The constant of integration is introduced here, and here only.
Check the integration by differentiating
$\frac{d}{dt}\left(\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}\right)=6\mathrm{e}^{3t}\cos{\left(t\right)}$
Differentiating the answer returns the integrand, so the integration is correct.
Divide through by the integrating factor
$I=\frac{\frac{3\left(\sin{\left(t\right)}+3\cos{\left(t\right)}\right)\mathrm{e}^{3t}}{5}+C}{\mathrm{e}^{3t}}$
This makes $I$ the subject and gives the general solution.
Apply the boundary condition
$I=0\ \text{when}\ t=0\ \Rightarrow\ 0=C+\frac{9}{5}$
Substituting the given values turns the general solution into an equation for $C$ .
Solve for the arbitrary constant
$C=-\frac{9}{5}$
This single value of $C$ selects the one curve through the given point.
Reject option B
$\frac{d}{dt}\left(\frac{2\sin{\left(t\right)}}{5}+\frac{6\cos{\left(t\right)}}{5}-\frac{6\mathrm{e}^{-3t}}{5}\right)\ \text{gives a residual of}\ -2\cos{\left(t\right)}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option C
$\frac{d}{dt}\left(\frac{9\sin{\left(t\right)}}{5}+\frac{3\cos{\left(t\right)}}{5}-\frac{3\mathrm{e}^{-3t}}{5}\right)\ \text{gives a residual of}\ \frac{24\sin{\left(t\right)}}{5}-\frac{12\cos{\left(t\right)}}{5}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option D
$\frac{d}{dt}\left(2\sin{\left(t\right)}-2\mathrm{e}^{-3t}\right)\ \text{gives a residual of}\ 6\sin{\left(t\right)}-4\cos{\left(t\right)}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Select the correct solution
$I=\frac{3\sin{\left(t\right)}}{5}+\frac{9\cos{\left(t\right)}}{5}-\frac{9\mathrm{e}^{-3t}}{5}$
This is the only option with a zero residual in the differential equation.

Answer

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

Question 2

9 markschallenging

Which of the following best describes the family of solution curves of the differential equation

\frac{dy}{dx}=-\frac{x}{4y}

Show worked solution

Worked solution

Check that the variables separate
$\frac{dy}{dx}=\left(-\frac{x}{4}\right)\times \frac{1}{y}$
The right-hand side is a function of $x$ multiplied by a function of $y$ , so the equation is separable.
Separate the variables
$\int y\,dy=\int -\frac{x}{4}\,dx$
Divide by $\frac{1}{y}$ and multiply by $dx$ , then integrate both sides.
Integrate both sides
$\frac{y^{2}}{2}=-\frac{x^{2}}{8}+c$
A single arbitrary constant is enough; it is placed on the right.
Make $y$ the subject
$y=\frac{\sqrt{C-x^{2}}}{2}$
Rearranging gives the general solution explicitly, with the relabelled arbitrary constant written as $C$ .
Reject option B
$\frac{d}{dx}\left(\sqrt{C-x^{2}}\right)\ \text{gives a residual of}\ -\frac{3x}{4\sqrt{C-x^{2}}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option C
$\frac{d}{dx}\left(\frac{C}{4}-\frac{x^{2}}{4}\right)\ \text{gives a residual of}\ -\frac{x}{2}+\frac{x}{C-x^{2}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option D
$\frac{d}{dx}\left(\frac{\sqrt{C+x^{2}}}{2}\right)\ \text{gives a residual of}\ \frac{x}{\sqrt{C+x^{2}}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option E
$\frac{d}{dx}\left(C-\frac{x}{4}\right)\ \text{gives a residual of}\ \frac{-C+\frac{5x}{4}}{4C-x}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Differentiate the solution as a check
$\frac{dy}{dx}=-\frac{x}{2\sqrt{C-x^{2}}}$
The derivative is needed to test the answer in the original equation.
Substitute the solution back into the equation
$\text{LHS}=-\frac{x}{2\sqrt{C-x^{2}}},\qquad\text{RHS}=-\frac{x}{2\sqrt{C-x^{2}}}$
The two sides agree identically, so the residual is $0$ and the solution is genuine.
Note the shape of the implicit solution
$y\ \text{is a function of}\ x\ \text{only, and vice versa}$
Separation works precisely because each side involves one variable only.
Interpret the arbitrary constant
$C\ \text{varies}\ \Rightarrow\ \text{a family of solution curves}$
Different values of $C$ give different curves, all satisfying the same equation.
Note that one curve passes through each point
$\left(x_{0},y_{0}\right)\ \Rightarrow\ \text{one value of}\ C$
A boundary condition therefore picks out exactly one member of the family.
Recall the standard linear form
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
A first-order equation is linear when $y$ and $\frac{dy}{dx}$ both appear to the first power only.
Select the correct description of the family
$y=\frac{\sqrt{C-x^{2}}}{2}$
Every member of this family satisfies the equation, and no other family does.

Answer

y=\frac{\sqrt{C-x^{2}}}{2}

Question 3

9 markschallenging

A metal block at

\theta^{\circ}\mathrm{C}

cools in a room whose temperature is constant;

t

is measured in minutes. The model gives the differential equation

\frac{d\theta}{dt}+\frac{\theta}{15}=\frac{4}{3}

. Given that

\theta=75

when

t=0

, which of the following best describes the behaviour of

\theta

t\to\infty

Show worked solution

Worked solution

Identify $P\left(t\right)$ and $Q\left(t\right)$
$P\left(t\right)=\frac{1}{15},\qquad Q\left(t\right)=\frac{4}{3}$
$P$ is the coefficient of $\theta$ once the equation is in standard form.
Integrate $P\left(t\right)$
$\int \frac{1}{15}\,dt=\frac{t}{15}$
No constant of integration is needed at this stage.
Form the integrating factor
$\mu=\mathrm{e}^{\int \frac{1}{15}\,dt}=\mathrm{e}^{\frac{t}{15}}$
This is the multiplier that makes the left-hand side exact.
Multiply the equation through by the integrating factor
$\mathrm{e}^{\frac{t}{15}}\frac{d\theta}{dt}+\frac{\mathrm{e}^{\frac{t}{15}}}{15}\theta=\frac{4\mathrm{e}^{\frac{t}{15}}}{3}$
Every term of the equation is multiplied, so the equation is unchanged.
Recognise the left-hand side as an exact derivative
$\frac{d}{dt}\left(\mathrm{e}^{\frac{t}{15}}\theta\right)=\frac{4\mathrm{e}^{\frac{t}{15}}}{3}$
By the product rule the left-hand side is exactly the derivative of $\mathrm{e}^{\frac{t}{15}}\theta$ , which is the whole point of the integrating factor.
Integrate both sides with respect to $t$
$\mathrm{e}^{\frac{t}{15}}\theta=\int \frac{4\mathrm{e}^{\frac{t}{15}}}{3}\,dt$
Integrating an exact derivative simply undoes it.
Carry out the integration
$\mathrm{e}^{\frac{t}{15}}\theta=\int \frac{4\mathrm{e}^{\frac{t}{15}}}{3}\,dt=20\mathrm{e}^{\frac{t}{15}}+C$
The constant of integration is introduced here, and here only.
Check the integration by differentiating
$\frac{d}{dt}\left(20\mathrm{e}^{\frac{t}{15}}\right)=\frac{4\mathrm{e}^{\frac{t}{15}}}{3}$
Differentiating the answer returns the integrand, so the integration is correct.
Divide through by the integrating factor
$\theta=\frac{20\mathrm{e}^{\frac{t}{15}}+C}{\mathrm{e}^{\frac{t}{15}}}$
This makes $\theta$ the subject and gives the general solution.
Apply the boundary condition
$\theta=75\ \text{when}\ t=0\ \Rightarrow\ 75=C+20$
Substituting the given values turns the general solution into an equation for $C$ .
Solve for the arbitrary constant
$C=55$
This single value of $C$ selects the one curve through the given point.
State the particular solution
$\theta=20+55\mathrm{e}^{-\frac{t}{15}}$
The behaviour of the model is read directly from this expression.
Split the solution into a steady part and a transient part
$\theta_{\text{steady}}=20,\qquad \theta_{\text{transient}}=55\mathrm{e}^{-\frac{t}{15}}$
The second term is the transient: it is the part that dies away.
Take the limit of the transient term
$55\mathrm{e}^{-\frac{t}{15}}\to 0\ \text{as}\ t\to\infty$
A negative exponential tends to zero, so only the steady part survives.
Differentiate the solution as a check
$\frac{d\theta}{dt}=-\frac{11\mathrm{e}^{-\frac{t}{15}}}{3}$
The derivative is needed to test the answer in the original equation.
Substitute the solution back into the equation
$\text{LHS}=\frac{4}{3},\qquad\text{RHS}=\frac{4}{3}$
The two sides agree identically, so the residual is $0$ and the solution is genuine.
State the limiting behaviour
$\theta\to 20\ \text{as}\ t\to\infty$
The model therefore settles down to this limiting value.

Answer

\theta\to 20

Question 4

9 markschallenging

Which of the following is the general solution of the differential equation

\frac{dy}{dx}+y\tan{\left(x\right)}=\sec{\left(x\right)}

Show worked solution

Worked solution

Identify $P\left(x\right)$ and $Q\left(x\right)$
$P\left(x\right)=\tan{\left(x\right)},\qquad Q\left(x\right)=\sec{\left(x\right)}$
$P$ is the coefficient of $y$ once the equation is in standard form.
Integrate $P\left(x\right)$
$\int \tan{\left(x\right)}\,dx=-\ln{\left(\cos{\left(x\right)}\right)}$
No constant of integration is needed at this stage.
Form the integrating factor
$\mu=\mathrm{e}^{\int \tan{\left(x\right)}\,dx}=\mathrm{e}^{-\ln{\left(\cos{\left(x\right)}\right)}}=\sec{\left(x\right)}$
This is the multiplier that makes the left-hand side exact.
Multiply the equation through by the integrating factor
$\sec{\left(x\right)}\frac{dy}{dx}+\tan{\left(x\right)}\sec{\left(x\right)}y=\sec^{2}{\left(x\right)}$
Every term of the equation is multiplied, so the equation is unchanged.
Recognise the left-hand side as an exact derivative
$\frac{d}{dx}\left(\sec{\left(x\right)}y\right)=\sec^{2}{\left(x\right)}$
By the product rule the left-hand side is exactly the derivative of $\sec{\left(x\right)}y$ , which is the whole point of the integrating factor.
Integrate both sides with respect to $x$
$\sec{\left(x\right)}y=\int \sec^{2}{\left(x\right)}\,dx$
Integrating an exact derivative simply undoes it.
Carry out the integration
$\sec{\left(x\right)}y=\int \sec^{2}{\left(x\right)}\,dx=\tan{\left(x\right)}+C$
The constant of integration is introduced here, and here only.
Check the integration by differentiating
$\frac{d}{dx}\left(\tan{\left(x\right)}\right)=\sec^{2}{\left(x\right)}$
Differentiating the answer returns the integrand, so the integration is correct.
Divide through by the integrating factor
$y=\frac{\tan{\left(x\right)}+C}{\sec{\left(x\right)}}$
This makes $y$ the subject and gives the general solution.
Tidy the general solution
$y=C\cos{\left(x\right)}+\sin{\left(x\right)}$
Splitting the fraction shows the particular part and the complementary part $\frac{C}{\sec{\left(x\right)}}$ separately.
Reject option B
$\frac{d}{dx}\left(C\sin{\left(x\right)}+\cos{\left(x\right)}\right)\ \text{gives a residual of}\ \frac{C-1}{\cos{\left(x\right)}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option C
$\frac{d}{dx}\left(C\sec{\left(x\right)}+\sin{\left(x\right)}\right)\ \text{gives a residual of}\ \frac{2C\sin{\left(x\right)}}{\cos^{2}{\left(x\right)}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option D
$\frac{d}{dx}\left(C\cos{\left(x\right)}+\tan{\left(x\right)}\right)\ \text{gives a residual of}\ 2\tan^{2}{\left(x\right)}-\sec{\left(x\right)}+1\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Reject option E
$\frac{d}{dx}\left(C\cos{\left(x\right)}-\sin{\left(x\right)}\right)\ \text{gives a residual of}\ -\frac{2}{\cos{\left(x\right)}}\neq 0$
Substituting this option into the differential equation leaves a non-zero residual, so it is not a solution.
Differentiate the solution as a check
$\frac{dy}{dx}=-C\sin{\left(x\right)}+\cos{\left(x\right)}$
The derivative is needed to test the answer in the original equation.
Select the correct solution
$y=C\cos{\left(x\right)}+\sin{\left(x\right)}$
This is the only option with a zero residual in the differential equation.

Answer

y=C\cos{\left(x\right)}+\sin{\left(x\right)}

Question 5

9 markschallenging

The differential equation

x\frac{dy}{dx}+2y=x^{3}

has a family of solution curves. Find the equation of the curve that passes through the point

\left(1,1\right)

Show worked solution

Worked solution

Divide through to reach the standard linear form
$x\frac{dy}{dx}+2y=x^{3}\quad\Rightarrow\quad \frac{dy}{dx}+\frac{2y}{x}=x^{2}$
The coefficient of $\frac{dy}{dx}$ must be $1$ before the integrating factor can be written down.
Identify $P\left(x\right)$ and $Q\left(x\right)$
$P\left(x\right)=\frac{2}{x},\qquad Q\left(x\right)=x^{2}$
$P$ is the coefficient of $y$ once the equation is in standard form.
Integrate $P\left(x\right)$
$\int \frac{2}{x}\,dx=2\ln{\left(x\right)}$
No constant of integration is needed at this stage.
Form the integrating factor
$\mu=\mathrm{e}^{\int \frac{2}{x}\,dx}=\mathrm{e}^{2\ln{\left(x\right)}}=x^{2}$
This is the multiplier that makes the left-hand side exact.
Multiply the equation through by the integrating factor
$x^{2}\frac{dy}{dx}+2xy=x^{4}$
Every term of the equation is multiplied, so the equation is unchanged.
Recognise the left-hand side as an exact derivative
$\frac{d}{dx}\left(x^{2}y\right)=x^{4}$
By the product rule the left-hand side is exactly the derivative of $x^{2}y$ , which is the whole point of the integrating factor.
Integrate both sides with respect to $x$
$x^{2}y=\int x^{4}\,dx$
Integrating an exact derivative simply undoes it.
Carry out the integration
$x^{2}y=\int x^{4}\,dx=\frac{x^{5}}{5}+C$
The constant of integration is introduced here, and here only.
Check the integration by differentiating
$\frac{d}{dx}\left(\frac{x^{5}}{5}\right)=x^{4}$
Differentiating the answer returns the integrand, so the integration is correct.
Divide through by the integrating factor
$y=\frac{\frac{x^{5}}{5}+C}{x^{2}}$
This makes $y$ the subject and gives the general solution.
Tidy the general solution
$y=\frac{C}{x^{2}}+\frac{x^{3}}{5}$
Splitting the fraction shows the particular part and the complementary part $\frac{C}{x^{2}}$ separately.
Apply the boundary condition
$y=1\ \text{when}\ x=1\ \Rightarrow\ 1=C+\frac{1}{5}$
Substituting the given values turns the general solution into an equation for $C$ .
Solve for the arbitrary constant
$C=\frac{4}{5}$
This single value of $C$ selects the one curve through the given point.
Differentiate the solution as a check
$\frac{dy}{dx}=\frac{3x^{5}-8}{5x^{3}}$
The derivative is needed to test the answer in the original equation.
State the equation of the curve
$y=\frac{x^{5}+4}{5x^{2}}$
This is the member of the family that passes through the given point.

Answer

y=\frac{x^{5}+4}{5x^{2}}

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