Further Maths Reducible differential equations Practice Questions

Free Further Maths Reducible differential equations practice questions with full step-by-step worked solutions. Covers euler-cauchy, second-order, substitution, auxiliary-equation. Practise exam-style problems and check your method.

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?
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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 dydx=(4x+y)2−4\frac{dy}{dx}=\left(4x+y\right)^{2}-4 is solved using the substitution z=4x+yz=4x+y. Which of the following is the general solution?
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Worked solution

  1. Substitute z=4x+yz=4x+y and differentiate

    dzdx=4+dydx\frac{dz}{dx}=4+\frac{dy}{dx}

    The right-hand side depends on xx and yy only through 4x+y4x+y.

  2. Reduce the equation to a separable equation in zz

    dzdx=z2\frac{dz}{dx}=z^{2}

    Replacing dydx\frac{dy}{dx} by dzdx−4\frac{dz}{dx}-4 removes xx and yy entirely.

  3. Separate, integrate and make zz the subject

    z=−1C+xz=-\frac{1}{C+x}

    The integral on the left is standard.

  4. Select the option that satisfies the differential equation

    y=−4x−1C+xy=-4x-\frac{1}{C+x}

    This is the general solution of the differential equation.

Answer
y=−4x−1C+xy=-4x-\frac{1}{C+x}
Question 3
4 marksintermediate
The differential equation dydx=(x+y)2−1\frac{dy}{dx}=\left(x+y\right)^{2}-1 is to be solved using the substitution z=x+yz=x+y. Given that y=−1y=-1 when x=0x=0, find yy in terms of xx.
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Worked solution

  1. Substitute z=x+yz=x+y and differentiate

    dzdx=1+dydx\frac{dz}{dx}=1+\frac{dy}{dx}

    The right-hand side depends on xx and yy only through x+yx+y.

  2. Reduce the equation to a separable equation in zz

    dzdx=z2\frac{dz}{dx}=z^{2}

    Replacing dydx\frac{dy}{dx} by dzdx−1\frac{dz}{dx}-1 removes xx and yy entirely.

  3. Separate, integrate and make zz the subject

    z=−1C+xz=-\frac{1}{C+x}

    The integral on the left is standard.

  4. Return to y=z−1xy=z-1x

    y=−x−1C+xy=-x-\frac{1}{C+x}

    Undoing the substitution gives the general solution.

  5. Apply the initial condition and solve for CC

    −1C=−1⇒C=1-\frac{1}{C}=-1\quad\Rightarrow\quad C=1

    The given point determines the arbitrary constant uniquely.

  6. Check the condition is satisfied

    y(0)=−1y\left(0\right)=-1

    Substituting the given value of xx reproduces the given value of yy.

  7. State the final answer

    y=−x−1x+1y=-x-\frac{1}{x+1}

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

Answer
y=−x−1x+1y=-x-\frac{1}{x+1}
Question 4
6 markshard
The differential equation xd2ydx2−2dydx=x3x\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}=x^{3}, where x>0x>0, is to be solved using the substitution u=dydxu=\frac{dy}{dx}. Find the general solution, giving yy in terms of xx and arbitrary constants CC and DD.
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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. 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.

  4. 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.

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

    xdudx−2u=x3x\frac{du}{dx}-2u=x^{3}

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

  6. Solve the first-order equation for uu

    u=x2(C+x)u=x^{2}\left(C+x\right)

    This introduces the first arbitrary constant.

  7. 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.

  8. Differentiate the answer once

    dydx=x2(C+x)\frac{dy}{dx}=x^{2}\left(C+x\right)

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

  9. Differentiate the answer a second time

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

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

  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 is a genuine solution.

  11. State the final answer

    y=Cx33+D+x44y=\frac{Cx^{3}}{3}+D+\frac{x^{4}}{4}

    This is the general solution of the differential equation.

Answer
y=Cx33+D+x44y=\frac{Cx^{3}}{3}+D+\frac{x^{4}}{4}
Question 5
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.
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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}

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