Further Maths Second-order differential equations Practice Questions

Free Further Maths Second-order differential equations practice questions with full step-by-step worked solutions. Covers auxiliary-equation, complementary-function, second-order, particular-integral. Practise exam-style problems and check your method.

auxiliary-equationcomplementary-functionsecond-orderparticular-integralinitial-conditionsboundary-conditions
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Write down the auxiliary equation for the differential equation d2ydx2−5dydx+6y=0\frac{d^{2}y}{dx^{2}}-5\frac{dy}{dx}+6y=0.
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Worked solution

  1. Try a solution of the form y=emxy=\mathrm{e}^{mx}

    y=emx,dydx=memx,d2ydx2=m2emxy=\mathrm{e}^{mx},\quad\frac{dy}{dx}=m\mathrm{e}^{mx},\quad\frac{d^{2}y}{dx^{2}}=m^{2}\mathrm{e}^{mx}

    Each derivative of emx\mathrm{e}^{mx} is a multiple of emx\mathrm{e}^{mx}.

  2. Substitute into the left-hand side

    (m2−5m+6)emx=0\left(m^{2}-5m+6\right)\mathrm{e}^{mx}=0

    Every term now carries the common factor emx\mathrm{e}^{mx}.

  3. State the auxiliary equation

    m2−5m+6=0m^{2}-5m+6=0

    This quadratic in mm is what the differential equation reduces to.

Answer
m2−5m+6=0m^{2}-5m+6=0
Question 2
2 markseasy
Write down the complementary function of the differential equation d2ydx2−4dydx+3y=6x\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+3y=6x.
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Worked solution

  1. Write down the auxiliary equation

    m2−4m+3=0m^{2}-4m+3=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.

  2. Solve the auxiliary equation

    m=1, m=3m=1,\ m=3

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

  3. Ignore the right-hand side

    d2ydx2−4dydx+3y=0\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+3y=0

    The complementary function is the general solution of the associated homogeneous equation.

  4. State the complementary function

    yc=Cex+De3xy_{c}=Ce^{x}+De^{3x}

    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}.

Answer
yc=Cex+De3xy_{c}=Ce^{x}+De^{3x}
Question 3
4 marksintermediate
Consider the differential equation d2ydx2−dydx−6y=0\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-6y=0. Given that y=3y=3 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. Write down the auxiliary equation

    m2−m−6=0m^{2}-m-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.

  2. Solve the auxiliary equation

    m=−2, m=3m=-2,\ m=3

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

  3. Write down the complementary function

    yc=Ce−2x+De3xy_{c}=Ce^{-2x}+De^{3x}

    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}.

  4. Write down the general solution

    y=Ce−2x+De3xy=Ce^{-2x}+De^{3x}

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

  5. Apply both conditions to the general solution

    y(0)=C+D=3,dydx∣x=0=−2C+3D=1y\left(0\right)=C+D=3,\quad \left.\frac{dy}{dx}\right|_{x=0}=-2C+3D=1

    Differentiating the general solution and substituting x=0x=0 gives two simultaneous equations.

  6. Solve for the arbitrary constants

    C=85,D=75C=\frac{8}{5},\quad D=\frac{7}{5}

    These are the only constants consistent with the conditions.

  7. Select the option that solves the differential equation

    y=7e3x5+8e−2x5y=\frac{7e^{3x}}{5}+\frac{8e^{-2x}}{5}

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

Answer
y=7e3x5+8e−2x5y=\frac{7e^{3x}}{5}+\frac{8e^{-2x}}{5}
Question 4
6 markshard
Consider the differential equation d2ydx2+y=0\frac{d^{2}y}{dx^{2}}+y=0. Given that y=3y=3 when x=0x=0 and y=4y=4 when x=π2x=\frac{\pi}{2}, 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=1a=1,\quad b=0,\quad c=1

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

  2. Write down the auxiliary equation

    m2+1=0m^{2}+1=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=i, m=−im=i,\ m=-i

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

  4. Write down the complementary function

    yc=Ccos⁡(x)+Dsin⁡(x)y_{c}=C\cos{\left(x\right)}+D\sin{\left(x\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. Write down the general solution

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

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

  6. Substitute the first boundary condition

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

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

  7. Substitute the second boundary condition

    y(π2)=D=4y\left(\frac{\pi}{2}\right)=D=4

    The value of yy at x=π2x=\frac{\pi}{2} gives the second equation for the constants.

  8. Solve the two equations for the arbitrary constants

    C=3,D=4C=3,\quad D=4

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

  9. Differentiate the answer twice

    d2ydx2=−4sin⁡(x)−3cos⁡(x)\frac{d^{2}y}{dx^{2}}=-4\sin{\left(x\right)}-3\cos{\left(x\right)}

    The second derivative is needed for the check.

  10. 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.

  11. State the final answer

    y=4sin⁡(x)+3cos⁡(x)y=4\sin{\left(x\right)}+3\cos{\left(x\right)}

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

Answer
y=4sin⁡(x)+3cos⁡(x)y=4\sin{\left(x\right)}+3\cos{\left(x\right)}
Question 5
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.
Show worked solution

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}

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