Hard Further Maths Second-order differential equations Questions

Challenging, exam-style Further Maths Second-order differential equations questions with worked solutions. Stretch yourself on the hardest auxiliary-equation, complementary-function, particular-integral, second-order problems.

auxiliary-equationcomplementary-functionparticular-integralsecond-orderresonanceinitial-conditions
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Consider the differential equation d2ydx2+6dydx+9y=4e−3x\frac{d^{2}y}{dx^{2}}+6\frac{dy}{dx}+9y=4e^{-3x}. Given that y=1y=1 and dydx=0\frac{dy}{dx}=0 when x=0x=0, find yy in terms of xx.
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Worked solution

  1. Identify the equation as linear with constant coefficients

    a=1,b=6,c=9a=1,\quad b=6,\quad c=9

    The coefficients are numbers, so the auxiliary-equation method applies.

  2. Write down the auxiliary equation

    m2+6m+9=0m^{2}+6m+9=0

    Putting y=emxy=\mathrm{e}^{mx} into the left-hand side and cancelling emx\mathrm{e}^{mx} turns the differential equation into this quadratic.

  3. Solve the auxiliary equation

    m=−3 (repeated)m=-3\ (\text{repeated})

    The discriminant is zero, so the auxiliary equation has one repeated root.

  4. Write down the complementary function

    yc=(C+Dx)e−3xy_{c}=\left(C+Dx\right)e^{-3x}

    A repeated root supplies only one exponential, so the second independent solution carries an extra factor of xx: yc=(C+Dx)emxy_{c}=\left(C+Dx\right)\mathrm{e}^{mx}.

  5. Look at the right-hand side

    RHS=4e−3x (exponential)\text{RHS}=4e^{-3x}\ \text{(exponential)}

    The shape of the right-hand side dictates the shape of the trial particular integral.

  6. Test the right-hand side against the complementary function

    4e−3x is already in yc4e^{-3x}\ \text{is already in}\ y_{c}

    The right-hand side is itself a term of the complementary function, so the usual trial function would give 00 on the left. Multiplying the trial function by an extra factor of xx repairs this.

  7. Choose the trial particular integral

    yp=λx2e−3xy_{p}=\lambda x^{2}e^{-3x}

    The trial function copies the shape of the right-hand side, with the extra factor of xx demanded by the clash with the complementary function.

  8. Compare coefficients on both sides

    2λ−4=02\lambda-4=0

    Powers of xx, exponentials, sines and cosines are independent, so their coefficients must match separately.

  9. Solve for the unknown coefficients

    λ=2\lambda=2

    These values are the only ones that make the trial function satisfy the equation.

  10. State the particular integral

    yp=2x2e−3xy_{p}=2x^{2}e^{-3x}

    Substituting the coefficients back into the trial function gives one solution of the full equation.

  11. Write down the general solution

    y=2x2e−3x+(C+Dx)e−3xy=2x^{2}e^{-3x}+\left(C+Dx\right)e^{-3x}

    The general solution of a linear equation is the complementary function plus the particular integral.

  12. Substitute the condition on yy

    y(0)=C=1y\left(0\right)=C=1

    This is the first equation satisfied by the two arbitrary constants.

  13. Differentiate the general solution

    dydx=(−3C−3Dx+D−6x2+4x)e−3x\frac{dy}{dx}=\left(-3C-3Dx+D-6x^{2}+4x\right)e^{-3x}

    The second condition involves the gradient, so the derivative of the general solution is needed.

  14. Substitute the condition on dydx\frac{dy}{dx}

    dydx∣x=0=−3C+D=0\left.\frac{dy}{dx}\right|_{x=0}=-3C+D=0

    This is the second equation satisfied by the two arbitrary constants.

  15. Solve the two equations for the arbitrary constants

    C=1,D=3C=1,\quad D=3

    Both conditions are now used, so the solution is completely determined.

  16. State the final answer

    y=2x2e−3x+3xe−3x+e−3xy=2x^{2}e^{-3x}+3xe^{-3x}+e^{-3x}

    This function satisfies the differential equation and both of the given conditions.

Answer
y=2x2e−3x+3xe−3x+e−3xy=2x^{2}e^{-3x}+3xe^{-3x}+e^{-3x}
Question 2
9 markschallenging
Use the substitution z=xyz=xy to solve the differential equation xd2ydx2+2dydx+4xy=8x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+4xy=8, where x>0x>0. Which of the following is the general solution?
Show worked solution

Worked solution

  1. Recall the structure of the general solution

    y=yc+ypy=y_{c}+y_{p}

    For a linear equation the general solution is the complementary function plus any one particular integral.

  2. State the substitution

    z=xyz=xy

    The equation is engineered so that this product is the new variable.

  3. Differentiate the substitution once

    dzdx=y+xdydx\frac{dz}{dx}=y+x\frac{dy}{dx}

    The product rule gives the first derivative of zz.

  4. Differentiate a second time

    d2zdx2=2dydx+xd2ydx2\frac{d^{2}z}{dx^{2}}=2\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}

    Differentiating y+xdydxy+x\frac{dy}{dx} again gives exactly the first two terms of the differential equation.

  5. Recognise the first two terms of the equation

    xd2ydx2+2dydx=d2zdx2x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=\frac{d^{2}z}{dx^{2}}

    This is why the substitution z=xyz=xy is the one suggested.

  6. Rewrite the whole equation in terms of zz

    d2zdx2+4z=8\frac{d^{2}z}{dx^{2}}+4z=8

    The remaining term is 4 xy=4 z4\,xy=4\,z, so the equation becomes a constant-coefficient equation in zz.

  7. Write down the auxiliary equation

    m2+4=0m^{2}+4=0

    The reduced equation is a standard constant-coefficient equation.

  8. Solve the auxiliary equation

    m=2i, m=−2im=2i,\ m=-2i

    The discriminant is negative, so the roots form a complex conjugate pair p±qip\pm qi.

  9. Write down the complementary function for zz

    zc=Ccos⁡(2x)+Dsin⁡(2x)z_{c}=C\cos{\left(2x\right)}+D\sin{\left(2x\right)}

    Roots p±qip\pm qi give the oscillatory complementary function yc=epx(Ccos⁡qx+Dsin⁡qx)y_{c}=\mathrm{e}^{px}\left(C\cos qx+D\sin qx\right).

  10. Choose a trial particular integral for zz

    yp=λy_{p}=\lambda

    The trial function copies the shape of the right-hand side 88.

  11. Compare coefficients and solve for them

    4λ−8=0 ⇒ λ=24\lambda-8=0\ \Rightarrow\ \lambda=2

    Matching coefficients determines the trial constants.

  12. Write down the general solution for zz

    z=Ccos⁡(2x)+Dsin⁡(2x)+2z=C\cos{\left(2x\right)}+D\sin{\left(2x\right)}+2

    This is the complementary function plus the particular integral.

  13. Differentiate the answer twice

    d2ydx2=2(Ccos⁡(2x)+Dsin⁡(2x)−2x2(Ccos⁡(2x)+Dsin⁡(2x))+2x(Csin⁡(2x)−Dcos⁡(2x))+2)x3\frac{d^{2}y}{dx^{2}}=\frac{2\left(C\cos{\left(2x\right)}+D\sin{\left(2x\right)}-2x^{2}\left(C\cos{\left(2x\right)}+D\sin{\left(2x\right)}\right)+2x\left(C\sin{\left(2x\right)}-D\cos{\left(2x\right)}\right)+2\right)}{x^{3}}

    The second derivative is needed for the check.

  14. Substitute the answer back into the differential equation

    LHS−RHS=0for all x\text{LHS}-\text{RHS}=0\quad\text{for all }x

    The residual reduces identically to zero, so the answer really does solve the equation.

  15. Select the option that solves the differential equation

    y=Ccos⁡(2x)+Dsin⁡(2x)+2xy=\frac{C\cos{\left(2x\right)}+D\sin{\left(2x\right)}+2}{x}

    This is the general solution: it contains the two arbitrary constants a second-order equation must have.

Answer
y=Ccos⁡(2x)+Dsin⁡(2x)+2xy=\frac{C\cos{\left(2x\right)}+D\sin{\left(2x\right)}+2}{x}
Question 3
9 markschallenging
Use the substitution x=eux=\mathrm{e}^{u} to solve the differential equation 2x2d2ydx2+3xdydx−y=02x^{2}\frac{d^{2}y}{dx^{2}}+3x\frac{dy}{dx}-y=0, where x>0x>0. Which of the following is the general solution?
Show worked solution

Worked solution

  1. Recall the structure of the general solution

    y=yc+ypy=y_{c}+y_{p}

    For a linear equation the general solution is the complementary function plus any one particular integral.

  2. Recall why the trial solution y=emxy=\mathrm{e}^{mx} works

    dydx=memx,d2ydx2=m2emx\frac{dy}{dx}=m\mathrm{e}^{mx},\qquad\frac{d^{2}y}{dx^{2}}=m^{2}\mathrm{e}^{mx}

    Every derivative of emx\mathrm{e}^{mx} is a multiple of emx\mathrm{e}^{mx}, which is what turns the differential equation into an algebraic one.

  3. Recall the auxiliary equation

    ad2ydx2+bdydx+cy=0 ⇒ am2+bm+c=0a\frac{d^{2}y}{dx^{2}}+b\frac{dy}{dx}+cy=0\ \Rightarrow\ am^{2}+bm+c=0

    Substituting y=emxy=\mathrm{e}^{mx} and cancelling the non-zero factor emx\mathrm{e}^{mx} leaves this quadratic.

  4. Recall the complementary function for two distinct real roots

    yc=Cem1x+Dem2xy_{c}=C\mathrm{e}^{m_{1}x}+D\mathrm{e}^{m_{2}x}

    Each root contributes one independent exponential solution.

  5. Recall the complementary function for a repeated root

    yc=(C+Dx)emxy_{c}=\left(C+Dx\right)\mathrm{e}^{mx}

    A repeated root gives only one exponential, so a factor of xx produces the second independent solution.

  6. Recall the complementary function for complex roots

    yc=epx(Ccos⁡qx+Dsin⁡qx)y_{c}=\mathrm{e}^{px}\left(C\cos qx+D\sin qx\right)

    Writing m=p±qim=p\pm qi and using Euler's relation turns the complex exponentials into a real oscillation with an exponential envelope.

  7. State the substitution and the link between uu and xx

    x=eu⇔u=ln⁡xx=\mathrm{e}^{u}\quad\Leftrightarrow\quad u=\ln x

    The substitution replaces powers of xx by exponentials in uu, which is what removes the variable coefficients.

  8. Differentiate once with the chain rule

    dydx=1xdydu⇒xdydx=dydu\frac{dy}{dx}=\frac{1}{x}\frac{dy}{du}\quad\Rightarrow\quad x\frac{dy}{dx}=\frac{dy}{du}

    Because dudx=1x\frac{du}{dx}=\frac{1}{x}, the factor of xx in the equation is absorbed exactly.

  9. Differentiate again to convert the second-derivative term

    x2d2ydx2=d2ydu2−dydux^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    Differentiating dydx=1xdydu\frac{dy}{dx}=\frac{1}{x}\frac{dy}{du} with the product rule and multiplying by x2x^{2} gives this standard result.

  10. Substitute both results into the equation and simplify

    2d2ydu2+dydu−y=02\frac{d^{2}y}{du^{2}}+\frac{dy}{du}-y=0

    Every variable coefficient has gone: the equation now has constant coefficients in uu.

  11. Write down the auxiliary equation of the reduced equation

    2m2+m−1=02m^{2}+m-1=0

    The reduced equation is a standard constant-coefficient equation.

  12. Solve the auxiliary equation

    m=−1, m=12m=-1,\ m=\frac{1}{2}

    The quadratic factorises, so the auxiliary equation has two distinct real roots.

  13. Write down the complementary function in terms of uu

    yc=Ce−u+Deu2y_{c}=Ce^{-u}+De^{\frac{u}{2}}

    Two distinct real roots give two independent exponential solutions, so the complementary function is yc=Cem1u+Dem2uy_{c}=C\mathrm{e}^{m_{1}u}+D\mathrm{e}^{m_{2}u}.

  14. Convert every term back to xx using emu=xm\mathrm{e}^{mu}=x^{m}

    e−u=1x,eu2=x\mathrm{e}^{-u}=\frac{1}{x},\quad \mathrm{e}^{\frac{u}{2}}=\sqrt{x}

    Since u=ln⁡xu=\ln x, each exponential in uu collapses to a power of xx; this is the step that undoes the substitution.

  15. Differentiate the answer twice

    d2ydx2=2Cx3−D4x32\frac{d^{2}y}{dx^{2}}=\frac{2C}{x^{3}}-\frac{D}{4x^{\frac{3}{2}}}

    The second derivative is needed for the check.

  16. Substitute the answer back into the differential equation

    LHS−RHS=0for all x\text{LHS}-\text{RHS}=0\quad\text{for all }x

    The residual reduces identically to zero, so the answer really does solve the equation.

  17. Select the option that solves the differential equation

    y=Cx+Dxy=\frac{C}{x}+D\sqrt{x}

    This is the general solution: it contains the two arbitrary constants a second-order equation must have.

Answer
y=Cx+Dxy=\frac{C}{x}+D\sqrt{x}
Question 4
9 markschallenging
Consider the differential equation d2ydx2+5dydx+6y=10e−x\frac{d^{2}y}{dx^{2}}+5\frac{dy}{dx}+6y=10e^{-x}. Given that y=1y=1 and dydx=1\frac{dy}{dx}=1 when x=0x=0, which of the following is the correct solution?
Show worked solution

Worked solution

  1. Identify the equation as linear with constant coefficients

    a=1,b=5,c=6a=1,\quad b=5,\quad c=6

    The coefficients are numbers, so the auxiliary-equation method applies.

  2. Write down the auxiliary equation

    m2+5m+6=0m^{2}+5m+6=0

    Putting y=emxy=\mathrm{e}^{mx} into the left-hand side and cancelling emx\mathrm{e}^{mx} turns the differential equation into this quadratic.

  3. Solve the auxiliary equation

    m=−3, m=−2m=-3,\ m=-2

    The quadratic factorises, so the auxiliary equation has two distinct real roots.

  4. Write down the complementary function

    yc=Ce−3x+De−2xy_{c}=Ce^{-3x}+De^{-2x}

    Two distinct real roots give two independent exponential solutions, so the complementary function is yc=Cem1x+Dem2xy_{c}=C\mathrm{e}^{m_{1}x}+D\mathrm{e}^{m_{2}x}.

  5. Look at the right-hand side

    RHS=10e−x (exponential)\text{RHS}=10e^{-x}\ \text{(exponential)}

    The shape of the right-hand side dictates the shape of the trial particular integral.

  6. Test the right-hand side against the complementary function

    10e−x is not in yc10e^{-x}\ \text{is not in}\ y_{c}

    The right-hand side is not part of the complementary function, so the trial function may be taken in the same shape as the right-hand side.

  7. Choose the trial particular integral

    yp=λe−xy_{p}=\lambda e^{-x}

    The trial function copies the shape of the right-hand side.

  8. Compare coefficients on both sides

    2λ−10=02\lambda-10=0

    Powers of xx, exponentials, sines and cosines are independent, so their coefficients must match separately.

  9. Solve for the unknown coefficients

    λ=5\lambda=5

    These values are the only ones that make the trial function satisfy the equation.

  10. State the particular integral

    yp=5e−xy_{p}=5e^{-x}

    Substituting the coefficients back into the trial function gives one solution of the full equation.

  11. Write down the general solution

    y=Ce−3x+De−2x+5e−xy=Ce^{-3x}+De^{-2x}+5e^{-x}

    The general solution of a linear equation is the complementary function plus the particular integral.

  12. Substitute the condition on yy

    y(0)=C+D+5=1y\left(0\right)=C+D+5=1

    This is the first equation satisfied by the two arbitrary constants.

  13. Differentiate the general solution

    dydx=(−3C−2Dex−5e2x)e−3x\frac{dy}{dx}=\left(-3C-2De^{x}-5e^{2x}\right)e^{-3x}

    The second condition involves the gradient, so the derivative of the general solution is needed.

  14. Substitute the condition on dydx\frac{dy}{dx}

    dydx∣x=0=−3C−2D−5=1\left.\frac{dy}{dx}\right|_{x=0}=-3C-2D-5=1

    This is the second equation satisfied by the two arbitrary constants.

  15. Solve the two equations for the arbitrary constants

    C=2,D=−6C=2,\quad D=-6

    Both conditions are now used, so the solution is completely determined.

  16. Select the option that solves the differential equation

    y=5e−x−6e−2x+2e−3xy=5e^{-x}-6e^{-2x}+2e^{-3x}

    This function satisfies the differential equation and both of the given conditions.

Answer
y=5e−x−6e−2x+2e−3xy=5e^{-x}-6e^{-2x}+2e^{-3x}
Question 5
9 markschallenging
Consider the differential equation d2ydx2+4y=8\frac{d^{2}y}{dx^{2}}+4y=8. Given that y=3y=3 when x=0x=0 and y=5y=5 when x=π4x=\frac{\pi}{4}, find yy in terms of xx.
Show worked solution

Worked solution

  1. Identify the equation as linear with constant coefficients

    a=1,b=0,c=4a=1,\quad b=0,\quad c=4

    The coefficients are numbers, so the auxiliary-equation method applies.

  2. Write down the auxiliary equation

    m2+4=0m^{2}+4=0

    Putting y=emxy=\mathrm{e}^{mx} into the left-hand side and cancelling emx\mathrm{e}^{mx} turns the differential equation into this quadratic.

  3. Solve the auxiliary equation

    m=2i, m=−2im=2i,\ m=-2i

    The discriminant is negative, so the roots form a complex conjugate pair p±qip\pm qi.

  4. Write down the complementary function

    yc=Ccos⁡(2x)+Dsin⁡(2x)y_{c}=C\cos{\left(2x\right)}+D\sin{\left(2x\right)}

    Roots p±qip\pm qi give the oscillatory complementary function yc=epx(Ccos⁡qx+Dsin⁡qx)y_{c}=\mathrm{e}^{px}\left(C\cos qx+D\sin qx\right).

  5. Look at the right-hand side

    RHS=8 (constant)\text{RHS}=8\ \text{(constant)}

    The shape of the right-hand side dictates the shape of the trial particular integral.

  6. Test the right-hand side against the complementary function

    8 is not in yc8\ \text{is not in}\ y_{c}

    The right-hand side is not part of the complementary function, so the trial function may be taken in the same shape as the right-hand side.

  7. Choose the trial particular integral

    yp=λy_{p}=\lambda

    The trial function copies the shape of the right-hand side.

  8. Compare coefficients on both sides

    4λ−8=04\lambda-8=0

    Powers of xx, exponentials, sines and cosines are independent, so their coefficients must match separately.

  9. Solve for the unknown coefficients

    λ=2\lambda=2

    These values are the only ones that make the trial function satisfy the equation.

  10. State the particular integral

    yp=2y_{p}=2

    Substituting the coefficients back into the trial function gives one solution of the full equation.

  11. Write down the general solution

    y=Ccos⁡(2x)+Dsin⁡(2x)+2y=C\cos{\left(2x\right)}+D\sin{\left(2x\right)}+2

    The general solution of a linear equation is the complementary function plus the particular integral.

  12. Substitute the first boundary condition

    y(0)=C+2=3y\left(0\right)=C+2=3

    The value of yy at x=0x=0 gives the first equation for the constants.

  13. Substitute the second boundary condition

    y(π4)=D+2=5y\left(\frac{\pi}{4}\right)=D+2=5

    The value of yy at x=π4x=\frac{\pi}{4} gives the second equation for the constants.

  14. Solve the two equations for the arbitrary constants

    C=1,D=3C=1,\quad D=3

    Both conditions are now used, so the solution is completely determined.

  15. State the final answer

    y=3sin⁡(2x)+cos⁡(2x)+2y=3\sin{\left(2x\right)}+\cos{\left(2x\right)}+2

    This function satisfies the differential equation and both of the given conditions.

Answer
y=3sin⁡(2x)+cos⁡(2x)+2y=3\sin{\left(2x\right)}+\cos{\left(2x\right)}+2

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