Second-order differential equations Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Second-order differential equations questions. See exactly how to solve problems on auxiliary-equation, complementary-function, second-order, particular-integral.

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.

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 auxiliary equation for the differential equation d2ydx2+2dydx+5y=0\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+5y=0.

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

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

  3. Cancel the non-zero factor emx\mathrm{e}^{mx}

    emx≠0 for all x\mathrm{e}^{mx}\neq0\ \text{for all }x

    An exponential is never zero, so it can be divided out safely.

  4. State the auxiliary equation

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

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

Answer
m2+2m+5=0m^{2}+2m+5=0
Question 3
2 markseasy
Write down the auxiliary equation for the differential equation d2ydx2−6dydx+9y=0\frac{d^{2}y}{dx^{2}}-6\frac{dy}{dx}+9y=0.

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

    (m2−6m+9)emx=0\left(m^{2}-6m+9\right)\mathrm{e}^{mx}=0

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

  3. State the auxiliary equation

    m2−6m+9=0m^{2}-6m+9=0

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

Answer
m2−6m+9=0m^{2}-6m+9=0
Question 4
2 markseasy
The differential equation d2ydx2−dydx−6y=0\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-6y=0 is to be solved. Which of the following gives the roots of the auxiliary equation?

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. Work out the discriminant

    b2−4ac=25b^{2}-4ac=25

    The sign of the discriminant decides which of the three cases applies.

  3. Apply the quadratic formula

    m=1±252m=\frac{1\pm\sqrt{25}}{2}

    The two roots come straight from the formula.

  4. Select the correct pair of roots

    m=−2, m=3m=-2,\ m=3

    These are the two roots of the auxiliary equation.

Answer
m=−2, m=3m=-2,\ m=3
Question 5
2 markseasy
The differential equation d2ydx2−4dydx+13y=0\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+13y=0 is to be solved. Which of the following gives the roots of the auxiliary equation?

Worked solution

  1. Write down the auxiliary equation

    m2−4m+13=0m^{2}-4m+13=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. Work out the discriminant

    b2−4ac=−36b^{2}-4ac=-36

    The sign of the discriminant decides which of the three cases applies.

  3. Select the correct pair of roots

    m=2+3i, m=2−3im=2+3i,\ m=2-3i

    These are the two roots of the auxiliary equation.

Answer
m=2+3i, m=2−3im=2+3i,\ m=2-3i

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