Reducible differential equations Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Reducible differential equations questions. See exactly how to solve problems on euler-cauchy, second-order, substitution, auxiliary-equation.

euler-cauchysecond-ordersubstitutionauxiliary-equationbernoullifirst-order
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The differential equation x2d2ydx2+xdydx−y=0x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-y=0, where x>0x>0, is solved using the substitution x=eux=\mathrm{e}^{u}. Which of the following is the general solution?

Worked solution

  1. Apply the substitution x=eux=\mathrm{e}^{u}

    xdydx=dydu,x2d2ydx2=d2ydu2−dydux\frac{dy}{dx}=\frac{dy}{du},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    These two standard results follow from the chain rule with u=ln⁡xu=\ln x.

  2. Reduce the equation and form the auxiliary equation

    d2ydu2−y=0⇒m2−1=0\frac{d^{2}y}{du^{2}}-y=0\quad\Rightarrow\quad m^{2}-1=0

    The equation now has constant coefficients, so the auxiliary equation applies.

  3. Solve the auxiliary equation and build the complementary function

    m=−1,m=1,yc=Cx+Dxm=-1,\quad m=1,\quad y_{c}=\frac{C}{x}+Dx

    Using emu=xm\mathrm{e}^{mu}=x^{m} writes each term directly as a power of xx.

  4. Select the option that satisfies the differential equation

    y=Cx+Dxy=\frac{C}{x}+Dx

    This is the general solution of the differential equation.

Answer
y=Cx+Dxy=\frac{C}{x}+Dx
Question 2
2 markseasy
The differential equation x2d2ydx2+2xdydx−2y=0x^{2}\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}-2y=0, where x>0x>0, is to be solved using the substitution x=eux=\mathrm{e}^{u}. Find the general solution, giving yy in terms of xx and arbitrary constants CC and DD.

Worked solution

  1. Apply the substitution x=eux=\mathrm{e}^{u}

    xdydx=dydu,x2d2ydx2=d2ydu2−dydux\frac{dy}{dx}=\frac{dy}{du},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    These two standard results follow from the chain rule with u=ln⁡xu=\ln x.

  2. Reduce the equation and form the auxiliary equation

    d2ydu2+dydu−2y=0⇒m2+m−2=0\frac{d^{2}y}{du^{2}}+\frac{dy}{du}-2y=0\quad\Rightarrow\quad m^{2}+m-2=0

    The equation now has constant coefficients, so the auxiliary equation applies.

  3. Solve the auxiliary equation and build the complementary function

    m=−2,m=1,yc=Cx2+Dxm=-2,\quad m=1,\quad y_{c}=\frac{C}{x^{2}}+Dx

    Using emu=xm\mathrm{e}^{mu}=x^{m} writes each term directly as a power of xx.

  4. State the final answer

    y=Cx2+Dxy=\frac{C}{x^{2}}+Dx

    This is the general solution of the differential equation.

Answer
y=Cx2+Dxy=\frac{C}{x^{2}}+Dx
Question 3
2 markseasy
The differential equation x2d2ydx2−2y=0x^{2}\frac{d^{2}y}{dx^{2}}-2y=0, where x>0x>0, is to be solved using the substitution x=eux=\mathrm{e}^{u}. Find the general solution, giving yy in terms of xx and arbitrary constants CC and DD.

Worked solution

  1. Apply the substitution x=eux=\mathrm{e}^{u}

    xdydx=dydu,x2d2ydx2=d2ydu2−dydux\frac{dy}{dx}=\frac{dy}{du},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    These two standard results follow from the chain rule with u=ln⁡xu=\ln x.

  2. Reduce the equation and form the auxiliary equation

    d2ydu2−dydu−2y=0⇒m2−m−2=0\frac{d^{2}y}{du^{2}}-\frac{dy}{du}-2y=0\quad\Rightarrow\quad m^{2}-m-2=0

    The equation now has constant coefficients, so the auxiliary equation applies.

  3. Solve the auxiliary equation and build the complementary function

    m=−1,m=2,yc=C+Dx3xm=-1,\quad m=2,\quad y_{c}=\frac{C+Dx^{3}}{x}

    Using emu=xm\mathrm{e}^{mu}=x^{m} writes each term directly as a power of xx.

  4. State the final answer

    y=C+Dx3xy=\frac{C+Dx^{3}}{x}

    This is the general solution of the differential equation.

Answer
y=C+Dx3xy=\frac{C+Dx^{3}}{x}
Question 4
2 markseasy
The differential equation x2d2ydx2−xdydx+y=0x^{2}\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+y=0, where x>0x>0, is to be solved using the substitution x=eux=\mathrm{e}^{u}. Find the general solution, giving yy in terms of xx and arbitrary constants CC and DD.

Worked solution

  1. Apply the substitution x=eux=\mathrm{e}^{u}

    xdydx=dydu,x2d2ydx2=d2ydu2−dydux\frac{dy}{dx}=\frac{dy}{du},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    These two standard results follow from the chain rule with u=ln⁡xu=\ln x.

  2. Reduce the equation and form the auxiliary equation

    d2ydu2−2dydu+y=0⇒m2−2m+1=0\frac{d^{2}y}{du^{2}}-2\frac{dy}{du}+y=0\quad\Rightarrow\quad m^{2}-2m+1=0

    The equation now has constant coefficients, so the auxiliary equation applies.

  3. Solve the auxiliary equation and build the complementary function

    m=1 (repeated),yc=x(C+Dln⁡(x))m=1\ (\text{repeated}),\quad y_{c}=x\left(C+D\ln{\left(x\right)}\right)

    Using emu=xm\mathrm{e}^{mu}=x^{m} writes each term directly as a power of xx.

  4. State the final answer

    y=x(C+Dln⁡(x))y=x\left(C+D\ln{\left(x\right)}\right)

    This is the general solution of the differential equation.

Answer
y=x(C+Dln⁡(x))y=x\left(C+D\ln{\left(x\right)}\right)
Question 5
2 markseasy
The differential equation x2d2ydx2+3xdydx+y=0x^{2}\frac{d^{2}y}{dx^{2}}+3x\frac{dy}{dx}+y=0, where x>0x>0, is solved using the substitution x=eux=\mathrm{e}^{u}. Which of the following is the general solution?

Worked solution

  1. Apply the substitution x=eux=\mathrm{e}^{u}

    xdydx=dydu,x2d2ydx2=d2ydu2−dydux\frac{dy}{dx}=\frac{dy}{du},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{du^{2}}-\frac{dy}{du}

    These two standard results follow from the chain rule with u=ln⁡xu=\ln x.

  2. Reduce the equation and form the auxiliary equation

    d2ydu2+2dydu+y=0⇒m2+2m+1=0\frac{d^{2}y}{du^{2}}+2\frac{dy}{du}+y=0\quad\Rightarrow\quad m^{2}+2m+1=0

    The equation now has constant coefficients, so the auxiliary equation applies.

  3. Solve the auxiliary equation and build the complementary function

    m=−1 (repeated),yc=C+Dln⁡(x)xm=-1\ (\text{repeated}),\quad y_{c}=\frac{C+D\ln{\left(x\right)}}{x}

    Using emu=xm\mathrm{e}^{mu}=x^{m} writes each term directly as a power of xx.

  4. Select the option that satisfies the differential equation

    y=C+Dln⁡(x)xy=\frac{C+D\ln{\left(x\right)}}{x}

    This is the general solution of the differential equation.

Answer
y=C+Dln⁡(x)xy=\frac{C+D\ln{\left(x\right)}}{x}

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