Hard Further Maths Reducible differential equations Questions

Challenging, exam-style Further Maths Reducible differential equations questions with worked solutions. Stretch yourself on the hardest euler-cauchy, second-order, substitution, auxiliary-equation problems.

euler-cauchysecond-ordersubstitutionauxiliary-equationbernoullifirst-order
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The differential equation yd2ydx2+(dydx)2=12x2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=12x^{2} is to be solved using the substitution z=y2z=y^{2}. Given that y=1y=1 and dydx=1\frac{dy}{dx}=1 when x=0x=0, find yy in terms of xx.
Show worked solution

Worked solution

  1. Differentiate z=y2z=y^{2} once

    dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx}

    The chain rule gives the first derivative of zz.

  2. Differentiate z=y2z=y^{2} a second time

    d2zdx2=2yd2ydx2+2(dydx)2\frac{d^{2}z}{dx^{2}}=2y\frac{d^{2}y}{dx^{2}}+2\left(\frac{dy}{dx}\right)^{2}

    The product rule applied to 2ydydx2y\frac{dy}{dx} gives exactly twice the left-hand side.

  3. Recognise the left-hand side of the equation

    yd2ydx2+(dydx)2=12d2zdx2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=\frac{1}{2}\frac{d^{2}z}{dx^{2}}

    The whole left-hand side is half of d2zdx2\frac{d^{2}z}{dx^{2}}.

  4. Rewrite the differential equation in terms of zz

    d2zdx2=24x2\frac{d^{2}z}{dx^{2}}=24x^{2}

    The equation for zz can now be integrated directly.

  5. Integrate once with respect to xx

    dzdx=8x3+C\frac{dz}{dx}=8x^{3}+C

    The first integration introduces the first arbitrary constant.

  6. Integrate a second time

    z=Cx+D+2x4z=Cx+D+2x^{4}

    The second integration introduces the second arbitrary constant.

  7. Convert the condition on yy into a condition on zz

    z(0)=y2=1z\left(0\right)=y^{2}=1

    Since z=y2z=y^{2}, the value of yy gives the value of zz.

  8. Convert the condition on dydx\frac{dy}{dx} into a condition on dzdx\frac{dz}{dx}

    dzdx∣x=0=2ydydx=2\left.\frac{dz}{dx}\right|_{x=0}=2y\frac{dy}{dx}=2

    Using dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx} converts the gradient condition.

  9. Solve for both constants

    C=2,D=1C=2,\quad D=1

    Both conditions have now been used.

  10. Write down zz and take the square root

    z=2x4+2x+1⇒y=2x4+2x+1z=2x^{4}+2x+1\quad\Rightarrow\quad y=\sqrt{2x^{4}+2x+1}

    The initial condition selects the positive square root.

  11. Differentiate the answer once

    dydx=4x3+12x4+2x+1\frac{dy}{dx}=\frac{4x^{3}+1}{\sqrt{2x^{4}+2x+1}}

    The first derivative of the solution is needed to check the equation.

  12. Differentiate the answer a second time

    d2ydx2=8x6+16x3+12x2−1(2x4+2x+1)32\frac{d^{2}y}{dx^{2}}=\frac{8x^{6}+16x^{3}+12x^{2}-1}{\left(2x^{4}+2x+1\right)^{\frac{3}{2}}}

    The second derivative is needed for a second-order equation.

  13. 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 is a genuine solution.

  14. Check the initial condition

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

    The particular solution must reproduce the given value of yy.

  15. State the final answer

    y=2x4+2x+1y=\sqrt{2x^{4}+2x+1}

    This particular solution satisfies both the differential equation and the given conditions.

Answer
y=2x4+2x+1y=\sqrt{2x^{4}+2x+1}
Question 2
9 markschallenging
The differential equation yd2ydx2+(dydx)2=6xy\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=6x is to be solved using the substitution z=y2z=y^{2}. 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. Recall the chain rule used by every substitution

    dydx=dydu×dudx\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}

    A substitution changes the independent (or dependent) variable, and the chain rule links the derivatives.

  2. Recall the integrating-factor method for a linear equation

    dzdx+P(x)z=Q(x)⇒I=e∫P dx\frac{dz}{dx}+P(x)z=Q(x)\quad\Rightarrow\quad I=\mathrm{e}^{\int P\,dx}

    Once an equation is linear in the new variable it can be solved with an integrating factor.

  3. Differentiate z=y2z=y^{2} once

    dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx}

    The chain rule gives the first derivative of zz.

  4. Differentiate z=y2z=y^{2} a second time

    d2zdx2=2yd2ydx2+2(dydx)2\frac{d^{2}z}{dx^{2}}=2y\frac{d^{2}y}{dx^{2}}+2\left(\frac{dy}{dx}\right)^{2}

    The product rule applied to 2ydydx2y\frac{dy}{dx} gives exactly twice the left-hand side.

  5. Recognise the left-hand side of the equation

    yd2ydx2+(dydx)2=12d2zdx2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=\frac{1}{2}\frac{d^{2}z}{dx^{2}}

    The whole left-hand side is half of d2zdx2\frac{d^{2}z}{dx^{2}}.

  6. Rewrite the differential equation in terms of zz

    d2zdx2=12x\frac{d^{2}z}{dx^{2}}=12x

    The equation for zz can now be integrated directly.

  7. Integrate once with respect to xx

    dzdx=6x2+C\frac{dz}{dx}=6x^{2}+C

    The first integration introduces the first arbitrary constant.

  8. Integrate a second time

    z=Cx+D+2x3z=Cx+D+2x^{3}

    The second integration introduces the second arbitrary constant.

  9. Convert the condition on yy into a condition on zz

    z(0)=y2=1z\left(0\right)=y^{2}=1

    Since z=y2z=y^{2}, the value of yy gives the value of zz.

  10. Convert the condition on dydx\frac{dy}{dx} into a condition on dzdx\frac{dz}{dx}

    dzdx∣x=0=2ydydx=0\left.\frac{dz}{dx}\right|_{x=0}=2y\frac{dy}{dx}=0

    Using dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx} converts the gradient condition.

  11. Solve for both constants

    C=0,D=1C=0,\quad D=1

    Both conditions have now been used.

  12. Write down zz and take the square root

    z=2x3+1⇒y=2x3+1z=2x^{3}+1\quad\Rightarrow\quad y=\sqrt{2x^{3}+1}

    The initial condition selects the positive square root.

  13. Differentiate the answer once

    dydx=3x22x3+1\frac{dy}{dx}=\frac{3x^{2}}{\sqrt{2x^{3}+1}}

    The first derivative of the solution is needed to check the equation.

  14. Differentiate the answer a second time

    d2ydx2=3x(x3+2)(2x3+1)32\frac{d^{2}y}{dx^{2}}=\frac{3x\left(x^{3}+2\right)}{\left(2x^{3}+1\right)^{\frac{3}{2}}}

    The second derivative is needed for a second-order equation.

  15. 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 is a genuine solution.

  16. Check the initial condition

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

    The particular solution must reproduce the given value of yy.

  17. Check the condition on the derivative

    dydx∣x=0=0\left.\frac{dy}{dx}\right|_{x=0}=0

    The particular solution must also reproduce the given gradient.

  18. State the final answer

    y=2x3+1y=\sqrt{2x^{3}+1}

    This particular solution satisfies both the differential equation and the given conditions.

Answer
y=2x3+1y=\sqrt{2x^{3}+1}
Question 3
9 markschallenging
The differential equation yd2ydx2+(dydx)2=4y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=4 is to be solved using the substitution z=y2z=y^{2}. Given that y=2y=2 and dydx=1\frac{dy}{dx}=1 when x=0x=0, find yy in terms of xx.
Show worked solution

Worked solution

  1. Recall the chain rule used by every substitution

    dydx=dydu×dudx\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}

    A substitution changes the independent (or dependent) variable, and the chain rule links the derivatives.

  2. Differentiate z=y2z=y^{2} once

    dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx}

    The chain rule gives the first derivative of zz.

  3. Differentiate z=y2z=y^{2} a second time

    d2zdx2=2yd2ydx2+2(dydx)2\frac{d^{2}z}{dx^{2}}=2y\frac{d^{2}y}{dx^{2}}+2\left(\frac{dy}{dx}\right)^{2}

    The product rule applied to 2ydydx2y\frac{dy}{dx} gives exactly twice the left-hand side.

  4. Recognise the left-hand side of the equation

    yd2ydx2+(dydx)2=12d2zdx2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=\frac{1}{2}\frac{d^{2}z}{dx^{2}}

    The whole left-hand side is half of d2zdx2\frac{d^{2}z}{dx^{2}}.

  5. Rewrite the differential equation in terms of zz

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

    The equation for zz can now be integrated directly.

  6. Integrate once with respect to xx

    dzdx=8x+C\frac{dz}{dx}=8x+C

    The first integration introduces the first arbitrary constant.

  7. Integrate a second time

    z=Cx+D+4x2z=Cx+D+4x^{2}

    The second integration introduces the second arbitrary constant.

  8. Convert the condition on yy into a condition on zz

    z(0)=y2=4z\left(0\right)=y^{2}=4

    Since z=y2z=y^{2}, the value of yy gives the value of zz.

  9. Convert the condition on dydx\frac{dy}{dx} into a condition on dzdx\frac{dz}{dx}

    dzdx∣x=0=2ydydx=4\left.\frac{dz}{dx}\right|_{x=0}=2y\frac{dy}{dx}=4

    Using dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx} converts the gradient condition.

  10. Solve for both constants

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

    Both conditions have now been used.

  11. Write down zz and take the square root

    z=4x2+4x+4⇒y=2x2+x+1z=4x^{2}+4x+4\quad\Rightarrow\quad y=2\sqrt{x^{2}+x+1}

    The initial condition selects the positive square root.

  12. Differentiate the answer once

    dydx=2x+1x2+x+1\frac{dy}{dx}=\frac{2x+1}{\sqrt{x^{2}+x+1}}

    The first derivative of the solution is needed to check the equation.

  13. Differentiate the answer a second time

    d2ydx2=32(x2+x+1)32\frac{d^{2}y}{dx^{2}}=\frac{3}{2\left(x^{2}+x+1\right)^{\frac{3}{2}}}

    The second derivative is needed for a second-order equation.

  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 is a genuine solution.

  15. Check the initial condition

    y(0)=2y\left(0\right)=2

    The particular solution must reproduce the given value of yy.

  16. Check the condition on the derivative

    dydx∣x=0=1\left.\frac{dy}{dx}\right|_{x=0}=1

    The particular solution must also reproduce the given gradient.

  17. State the final answer

    y=2x2+x+1y=2\sqrt{x^{2}+x+1}

    This particular solution satisfies both the differential equation and the given conditions.

Answer
y=2x2+x+1y=2\sqrt{x^{2}+x+1}
Question 4
9 markschallenging
The differential equation yd2ydx2+(dydx)2=2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=2 is to be solved using the substitution z=y2z=y^{2}. 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. Differentiate z=y2z=y^{2} once

    dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx}

    The chain rule gives the first derivative of zz.

  2. Differentiate z=y2z=y^{2} a second time

    d2zdx2=2yd2ydx2+2(dydx)2\frac{d^{2}z}{dx^{2}}=2y\frac{d^{2}y}{dx^{2}}+2\left(\frac{dy}{dx}\right)^{2}

    The product rule applied to 2ydydx2y\frac{dy}{dx} gives exactly twice the left-hand side.

  3. Recognise the left-hand side of the equation

    yd2ydx2+(dydx)2=12d2zdx2y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=\frac{1}{2}\frac{d^{2}z}{dx^{2}}

    The whole left-hand side is half of d2zdx2\frac{d^{2}z}{dx^{2}}.

  4. Rewrite the differential equation in terms of zz

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

    The equation for zz can now be integrated directly.

  5. Integrate once with respect to xx

    dzdx=4x+C\frac{dz}{dx}=4x+C

    The first integration introduces the first arbitrary constant.

  6. Integrate a second time

    z=Cx+D+2x2z=Cx+D+2x^{2}

    The second integration introduces the second arbitrary constant.

  7. Convert the condition on yy into a condition on zz

    z(0)=y2=1z\left(0\right)=y^{2}=1

    Since z=y2z=y^{2}, the value of yy gives the value of zz.

  8. Convert the condition on dydx\frac{dy}{dx} into a condition on dzdx\frac{dz}{dx}

    dzdx∣x=0=2ydydx=0\left.\frac{dz}{dx}\right|_{x=0}=2y\frac{dy}{dx}=0

    Using dzdx=2ydydx\frac{dz}{dx}=2y\frac{dy}{dx} converts the gradient condition.

  9. Solve for both constants

    C=0,D=1C=0,\quad D=1

    Both conditions have now been used.

  10. Write down zz and take the square root

    z=2x2+1⇒y=2x2+1z=2x^{2}+1\quad\Rightarrow\quad y=\sqrt{2x^{2}+1}

    The initial condition selects the positive square root.

  11. Differentiate the answer once

    dydx=2x2x2+1\frac{dy}{dx}=\frac{2x}{\sqrt{2x^{2}+1}}

    The first derivative of the solution is needed to check the equation.

  12. Differentiate the answer a second time

    d2ydx2=2(2x2+1)32\frac{d^{2}y}{dx^{2}}=\frac{2}{\left(2x^{2}+1\right)^{\frac{3}{2}}}

    The second derivative is needed for a second-order equation.

  13. 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 is a genuine solution.

  14. Check the initial condition

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

    The particular solution must reproduce the given value of yy.

  15. Check the condition on the derivative

    dydx∣x=0=0\left.\frac{dy}{dx}\right|_{x=0}=0

    The particular solution must also reproduce the given gradient.

  16. State the final answer

    y=2x2+1y=\sqrt{2x^{2}+1}

    This particular solution satisfies both the differential equation and the given conditions.

Answer
y=2x2+1y=\sqrt{2x^{2}+1}
Question 5
9 markschallenging
The differential equation d2ydx2=2(dydx)2\frac{d^{2}y}{dx^{2}}=2\left(\frac{dy}{dx}\right)^{2} is solved using the substitution u=dydxu=\frac{dy}{dx}. Given that y=0y=0 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. Note that yy itself does not appear in the equation

    only dydx and d2ydx2 occur\text{only }\frac{dy}{dx}\text{ and }\frac{d^{2}y}{dx^{2}}\text{ occur}

    This is exactly when the order of the equation can be reduced.

  2. State the substitution

    u=dydx⇒dudx=d2ydx2u=\frac{dy}{dx}\quad\Rightarrow\quad\frac{du}{dx}=\frac{d^{2}y}{dx^{2}}

    The second-order equation becomes a first-order equation in uu.

  3. Rewrite the equation as a first-order equation in uu

    dudx=2u2\frac{du}{dx}=2u^{2}

    Only uu and dudx\frac{du}{dx} now appear.

  4. Solve the first-order equation for uu

    u=−1C+2xu=-\frac{1}{C+2x}

    This introduces the first arbitrary constant.

  5. Recall that u=dydxu=\frac{dy}{dx}, so yy is found by integrating uu

    y=∫u dxy=\int u\,dx

    One further integration recovers yy.

  6. Integrate uu with respect to xx

    y=D−ln⁡(C+2x)2y=D-\frac{\ln{\left(C+2x\right)}}{2}

    The second integration introduces the second arbitrary constant.

  7. Apply the condition on dydx\frac{dy}{dx}

    −1C=−1-\frac{1}{C}=-1

    The gradient condition fixes the constant introduced when solving for uu.

  8. Apply the condition on yy

    D−ln⁡(C)2=0D-\frac{\ln{\left(C\right)}}{2}=0

    The value condition fixes the constant of the second integration.

  9. Solve for both constants

    C=1,D=0C=1,\quad D=0

    Both conditions have now been used.

  10. Differentiate the answer once

    dydx=−12x+1\frac{dy}{dx}=-\frac{1}{2x+1}

    The first derivative of the solution is needed to check the equation.

  11. Differentiate the answer a second time

    d2ydx2=2(2x+1)2\frac{d^{2}y}{dx^{2}}=\frac{2}{\left(2x+1\right)^{2}}

    The second derivative is needed for a second-order equation.

  12. 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 is a genuine solution.

  13. Check the initial condition

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

    The particular solution must reproduce the given value of yy.

  14. Check the condition on the derivative

    dydx∣x=0=−1\left.\frac{dy}{dx}\right|_{x=0}=-1

    The particular solution must also reproduce the given gradient.

  15. Select the option that satisfies the differential equation

    y=−ln⁡(2x+1)2y=-\frac{\ln{\left(2x+1\right)}}{2}

    This particular solution satisfies both the differential equation and the given conditions.

Answer
y=−ln⁡(2x+1)2y=-\frac{\ln{\left(2x+1\right)}}{2}

Unlock 29 more Reducible differential equations questions

Create a free account to work through every Further Maths Reducible differential equations question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Reducible differential equations practice

Practice QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2Further complex numbersGroupsInequalitiesFurther integration techniques