University Differential Equations — Constant-Coefficient Linear Equations Worksheet

All 12 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Complex Numbers in Rectangular, Polar and Exponential Form

   Let $z=4-2i$ and $w=1+3i$. Write each of the following in the form $a+bi$ with $a$ and $b$ real.

   1. $zw$
   2. ${\displaystyle \frac{z}{w}}$
   3. $w^2-\bar{z}$, where $\bar{z}$ is the complex conjugate of $z$
2. Q2Complex Numbers in Rectangular, Polar and Exponential Form

   Throughout, the argument of a nonzero complex number is taken in $(-\pi ,\pi ]$.

   1. Write $z=-2\sqrt{3}+2i$ in polar form $r(\cos \theta +i\sin \theta )$ and in exponential form $r{e}^{i\theta }$.
   2. Write $6e^{-i\pi /4}$ in the form $a+bi$.
   3. Use Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$ to write $e^{(-1+2i)t}$, for real $t$, in the form $u(t)+iv(t)$ with $u$ and $v$ real-valued functions.
3. Q3Powers and Roots of a Complex Number

   Use polar or exponential form in each part.

   1. Compute $(1-i{)}^{10}$, giving the answer in the form $a+bi$.
   2. Find all three cube roots of $-8i$, each in the form $a+bi$.
4. Q4Superposition, Linear Independence and the Wronskian

   Here $y=y(x)$ and $\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$, $\cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$.

   1. Verify that $y_1=e^x$ and $y_2=\sinh x$ are both solutions of $y^{″}-y=0$ on $(-\infty ,\infty )$.
   2. Compute the Wronskian $W(y_1,y_2)$, and say what its value tells you about $y_1$ and $y_2$.
   3. Write the general solution of $y^{″}-y=0$ in terms of $y_1$ and $y_2$. Then write the solution $e^{-x}$ as a combination $c_1{y}_1+c_2{y}_2$.
5. Q5Distinct Real Roots of the Characteristic Equation

   Find the general solution of each equation, where $y=y(x)$.

   1. $y^{″}+y^{\prime }-6y=0$
   2. $2y^{″}-7y^{\prime }+3y=0$
   3. $y^{″}-5y^{\prime }=0$
6. Q6Distinct Real Roots of the Characteristic Equation

   Solve the initial value problem $y^{″}-y^{\prime }-2y=0,y(0)=2,y^{\prime }(0)=-5$, where $y=y(t)$. Then find the value of $t>0$ at which the solution is zero.

7. Q7Repeated Roots of the Characteristic Equation

   Find the general solution of each equation, where $y=y(x)$.

   1. $y^{″}+6y^{\prime }+9y=0$
   2. $4y^{″}-4y^{\prime }+y=0$
8. Q8Repeated Roots of the Characteristic Equation

   Solve the initial value problem $y^{″}-8y^{\prime }+16y=0,y(0)=3,y^{\prime }(0)=10$, where $y=y(t)$, and find where the solution crosses zero.

9. Q9Complex Conjugate Roots of the Characteristic Equation

   Find the general solution of each equation in real form, where $y=y(x)$.

   1. $y^{″}+2y^{\prime }+5y=0$
   2. $y^{″}+16y=0$
   3. $9y^{″}-6y^{\prime }+10y=0$
10. Q10Complex Conjugate Roots of the Characteristic Equation

    Solve the initial value problem $y^{″}-4y^{\prime }+13y=0,y(0)=1,y^{\prime }(0)=8$, where $y=y(t)$.

11. Q11Homogeneous Equations of Third and Higher Order

    Find the general solution of each equation in real form, where $y=y(x)$.

    1. $y^{‴}-3y^{″}-4y^{\prime }+12y=0$
    2. $y^{(4)}+8y^{″}+16y=0$
    3. $y^{‴}+8y=0$
12. Q12Synthesis — drawing on several topics in this unit

    Consider $y^{(4)}+16y=0$, where $y=y(x)$.

    1. Find the four fourth roots of $-16$, each in the form $a+bi$.
    2. Write the general solution of the equation in real form.
    3. Find the solution that satisfies $y(0)=1$ and $y^{\prime }(0)=0$ and stays bounded as $x\to \infty$.

The 8 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Complex numbers in rectangular, polar and exponential form

A complex root of a characteristic equation is only useful once you can move it between three forms: a + bi for adding and multiplying, r(cos θ + i sin θ) for seeing its size and direction, and re^iθ for raising it to a power. Q1 works entirely in rectangular form. A product is expanded like any binomial, with i² replaced by −1 at the end. A quotient needs one move you must remember rather than improvise.

Q2 moves between forms. For part (a), the modulus r is √(a² + b²), and the argument is where students lose marks: the calculator's arctangent only ever returns an angle in the first or fourth quadrant. Plot the point first, decide its quadrant from the signs of a and b, and then correct the reference angle so that θ lands in (−π, π] as the question requires. Part (b) runs the other way — evaluate the cosine and sine of the given angle and multiply by the modulus.

Powers and roots of a complex number

In exponential form, multiplying complex numbers multiplies their moduli and adds their arguments. That turns a tenth power, which would take nine multiplications in rectangular form, into two operations. Q3(a) asks for exactly that: write the base in polar form, raise the modulus to the power, multiply the argument by the power, and only then convert back to a + bi.

Q3(b) asks for all three cube roots. The number has one modulus and many possible arguments, each differing by a full turn, and dividing those arguments by three is what separates the roots. Write the argument with + 2πk, divide by three, and run k through 0, 1 and 2.

Superposition, linear independence and the Wronskian

For a homogeneous linear equation, any combination c₁y₁ + c₂y₂ of two solutions is again a solution. That is superposition, and it is why the whole chapter looks for just two solutions of a second-order equation. The catch is that the two must be genuinely different: neither may be a constant multiple of the other. The Wronskian is the test.

Q4(a) is verification: differentiate each function twice and substitute into the equation. The definition of sinh x is printed in the question so that its derivatives can be built from those of e^x and e^−x. Part (b) computes the Wronskian and asks what its value says. Simplify it completely before interpreting it.

Part (c) is the practical meaning of that conclusion. A fundamental set reaches every solution, including ones that look nothing like y₁ and y₂. To write a third solution as c₁y₁ + c₂y₂, expand the hyperbolic function back into exponentials and match the coefficients of e^x and of e^−x on both sides — two equations, two constants.

The characteristic equation, and the three cases

Try y = e^rx in ay″ + by′ + cy = 0. Every term carries the same exponential factor, which is never zero, so it cancels and leaves a polynomial in r. The derivatives become powers of r; the equation becomes algebra.

Distinct real roots

Two different real roots r₁ and r₂ give two exponentials, e^r₁x and e^r₂x, and their Wronskian is never zero, so they form a fundamental set. Q5 has three equations, and each one tests a different part of step 1 or 2. In part (b) the leading coefficient is not 1; keep it in the factoring rather than dividing it away and losing track of it. In part (c) the constant term is missing, which changes how the polynomial factors.

Q6 is an initial value problem in t. Write the general solution, differentiate it, and put t = 0 into both: that gives two linear equations in c₁ and c₂. The second part asks where the solution is zero for t > 0. Set the solution equal to zero, move one exponential to each side, and divide so that a single exponential remains; a logarithm then finishes it. Check that the t you find is positive, as the question demands.

Repeated roots

When the discriminant is zero, the characteristic equation gives only one exponential, e^rx, and one solution is not a general solution. The second solution is that exponential multiplied by x. Writing c₁e^rx + c₂e^rx is the classic error: those two terms are the same function, and their Wronskian says so.

Q7 asks for the general solution of two equations. Recognise the perfect square in each characteristic polynomial; in part (b), with a leading coefficient that is not 1, the square to look for is of a binomial with a coefficient in front of r.

Complex conjugate roots

A negative discriminant gives roots α ± βi. The two complex exponentials they produce are legitimate solutions, but the question wants the answer in real form. Q2(c) already did the conversion: by Euler's formula, the real and imaginary parts of e^{(α + βi)x} are e^αxcos βx and e^αxsin βx, and those two real functions are themselves solutions.

Q9 has three equations and each checks one reading of that formula. In part (a), complete the square or use the quadratic formula to find α and β. In part (b) there is no first-derivative term; work out what that does to the real part α, and what an exponential factor with that exponent contributes. In part (c) the leading coefficient is not 1; divide the whole quadratic formula by 2a, not just the square root.

Q10 is the complex case as an initial value problem. Put t = 0 into the general solution, where the sine vanishes and the cosine is 1, and one constant falls out at once. The derivative needs the product rule on both terms; evaluated at t = 0 it gives a second equation, which contains α as well as β. Leaving out the α contribution is the usual slip.

Homogeneous equations of third and higher order

Nothing new happens in theory: y⁽ⁿ⁾ becomes rⁿ, the characteristic polynomial has degree n, and each root contributes as before. A root of multiplicity k contributes e^rx, x e^rx, …, x^k−1e^rx, and a repeated complex pair gets the same powers of x in front of its sine and cosine. The work is all in the factoring. Q11 has three polynomials, each built to need a different factoring idea.

The synthesis question

Q12 takes one fourth-order equation through the whole set. Part (a) is Q3(b) with four roots instead of three: write the negative real number in exponential form, add 2πk to its argument, and divide by four. Part (b) groups those roots into conjugate pairs and applies the complex case to each pair, so the general solution has four terms and four constants. Part (c) adds a condition that is not an initial value: the solution must stay bounded as x → ∞. Decide, for each term, whether its exponential factor grows or decays as x increases; boundedness forces the constants on the growing terms to be zero, and the two initial conditions then fix the rest.

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Getting the most out of it

Name the case before you write a solution

After finding the roots, write one word — distinct, repeated or complex — next to them. The shape of the general solution follows from that word, and most wrong answers in this chapter are a correct root put into the wrong case's formula.

Plot every complex number you convert

A two-second sketch of the point in the plane tells you its quadrant, and the quadrant is what the arctangent cannot. It also shows at a glance whether a set of n-th roots is evenly spaced around the circle, as it must be.

Check a root by substituting it

Put each root back into the characteristic polynomial, or raise a complex root to the required power. It takes a line and catches the sign and fraction errors that otherwise propagate into every constant downstream.

Differentiate the general solution before using initial conditions

Write y and y′ out in full, with the product rule on every term that has one, before substituting t = 0. Substituting too early, or differentiating only the obvious part of a product, is how an initial value problem goes wrong after the hard part was right.

Common questions