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← Ordinary Differential Equations levels

Ordinary Differential Equations / LEVEL 1 · DIFFICULTY 4/5

Ordinary Differential Equations

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE UODE 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

First-Order Equations & Initial Values

Useful preparation: Integrals & the Fundamental Theorem

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For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A differential equation specifies how a quantity changes at each state, rather than giving the quantity directly. An initial value selects a trajectory from a family. Separation reverses a chain rule; an integrating factor reverses a product rule. Choosing a method means recognizing the structure that makes one of those reversals possible.

Initial-value problem
An equation y prime=f(t,y), together with y (t 0)=y 0\displaystyle y\, \left(t\, 0\right) = y\, 0, asks for a differentiable function satisfying both conditions on an interval containing t0.
Separable equation
An equation y prime=g(t)h(y) separates its dependence on time and state. Where h(y) is nonzero, integrating 1h (y)\displaystyle \frac{1}{h}\, \left(y\right) with respect to y can recover the trajectory.
Linear first-order equation
The form y prime+p (t) y=q (t)\displaystyle p\, \left(t\right)\, y = q\, \left(t\right) is linear in y and its derivative. An integrating factor mu(t)=exp⁡\displaystyle \operatorname{exp}(integral p(t) dt) turns the left side into one product derivative.

Conditions before calculation

  • Before dividing by h(y), record all equilibrium solutions h (y)=0\displaystyle h\, \left(y\right) = 0; otherwise valid constant solutions can disappear.
  • Continuous p and q on an interval give a unique solution of a linear initial-value problem throughout that interval. A leading coefficient must be nonzero before dividing into standard form.
  • For nonlinear problems, continuity of f ensures local existence; local Lipschitz continuity in y ensures local uniqueness. Even smooth equations may have solutions that blow up in finite time.

Why the method works

1. Inspect before integrating

Determine whether the equation separates or is linear. Some equations have both structures; others have neither. Identify the independent variable and units before manipulating symbols.

2. Reverse a differentiation rule

For a linear equation choose mu so that mu prime=p mu. The product rule then supplies a directly integrable derivative.

μ(t)=e∫p(t) dt,(μy)′=μq\mu(t)=e^{\int p(t)\,dt},\qquad (\mu y)'=\mu q

3. Use the initial condition and verify

Integrate, solve for the constant, then check the original derivative equation. Report the interval containing the initial time on which the result remains defined.

Worked examples

EXAMPLE 1

Solve y prime=2t y with y (0)=3\displaystyle y\, \left(0\right) = 3.

  1. The initial value is nonzero. Separating on a nonzero branch gives the logarithm of the absolute value.

    dyy=2t dt,ln⁡∣y∣=t2+C\frac{dy}{y}=2t\,dt,\qquad \ln|y|=t^2+C

  2. Exponentiation gives y=K exp⁡\displaystyle y = K\, \operatorname{exp}(t squared); the initial condition fixes K=3\displaystyle K = 3.

    y(t)=3et2y(t)=3e^{t^2}

  3. Differentiate the result and evaluate at zero to verify both requirements.

    y′=6tet2=2ty,y(0)=3y'=6te^{t^2}=2ty,\qquad y(0)=3

Conclusion: This solution is defined for all real t. The separate equilibrium y=0\displaystyle y = 0 belongs to the differential equation but fails this initial condition.

EXAMPLE 2

Solve y prime+2 y=4\displaystyle 2\, y = 4 with y (0)=1\displaystyle y\, \left(0\right) = 1.

  1. Choose the integrating factor exp⁡ (2 t)\displaystyle \operatorname{exp}\, \left(2\, t\right), whose derivative supplies the coefficient 2.

    (e2ty)′=4e2t(e^{2t}y)'=4e^{2t}

  2. Integrate and divide by the positive factor.

    e2ty=2e2t+C,y=2+Ce−2te^{2t}y=2e^{2t}+C,\qquad y=2+Ce^{-2t}

  3. The initial value gives C=−1\displaystyle C = -1; substituting the derivative verifies the equation.

    y=2−e−2t,y′+2y=2e−2t+4−2e−2t=4y=2-e^{-2t},\qquad y'+2y=2e^{-2t}+4-2e^{-2t}=4

Conclusion: The solution tends to the equilibrium 2 as time increases.

Pitfalls & counterexamples

Tempting claim: A smooth differential equation guarantees a solution for all time.

Correction: Smoothness gives local control; the solution can still become unbounded.

Test the claim: For y prime=y squared and y (0)=1\displaystyle y\, \left(0\right) = 1, the solution is 11−t\displaystyle \frac{1}{1 - t}, whose maximal interval containing zero is t<1\displaystyle t < 1.

Check your understanding

Does dividing y prime=y (1−y)\displaystyle y\, \left(1 - y\right) by y (1−y)\displaystyle y\, \left(1 - y\right) preserve every solution?

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Compare your reasoning

No.

The constant solutions y=0\displaystyle y = 0 and y=1\displaystyle y = 1 must be recorded separately before division.

Connect this idea

The same product-rule idea underlies conservation laws and Laplace-transform solutions. Qualitative slope information can check whether a computed trajectory rises, falls, or approaches an equilibrium.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    Solve for the constant y′=2 y\displaystyle y' = 2\, y, y (0)=3\displaystyle y\, \left(0\right) = 3. In y=C e2 x\displaystyle y = C\, e^{2\, x}, find C.

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  2. Problem 2

    An equation is y′+2 y=0\displaystyle y' + 2\, y = 0. If an integrating factor is eax\displaystyle e^{ax}, find a.

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  3. Problem 3

    An initial-value problem has y′=2 x\displaystyle y' = 2\, x and y (0)=4\displaystyle y\, \left(0\right) = 4. Find y(2).

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  4. Problem 4

    For y′=−2 (y−5)\displaystyle y' = -2\, \left(y - 5\right), find the equilibrium value of y.

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  5. Problem 5New context

    A growth model satisfies y′=2 y\displaystyle y' = 2\, y, y (0)=4\displaystyle y\, \left(0\right) = 4. In y=C e2 x\displaystyle y = C\, e^{2\, x}, find C.

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  6. Problem 6New context

    A cooling model is y′+3 y=0\displaystyle y' + 3\, y = 0. If an integrating factor is eax\displaystyle e^{ax}, find a.

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  7. Problem 7New context

    A motion model has y′=3 x\displaystyle y' = 3\, x and y (0)=4\displaystyle y\, \left(0\right) = 4. Find y(2).

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  8. Problem 8New context

    A stable adjustment follows y′=−3 (y−6)\displaystyle y' = -3\, \left(y - 6\right), find the equilibrium value of y.

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STAGE UODE 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Second-Order Equations & Linear Systems

Useful preparation: First-Order Equations & Initial Values

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A second-order equation tracks a state and its rate of change. Two initial numbers are needed because position alone does not determine motion. For constant coefficients, exponentials reproduce their own shape under differentiation, making them natural building blocks. The characteristic roots identify whether these building blocks grow, decay, or oscillate.

Second-order linear equation
The equation a y double prime+b y prime+c y=g (t)\displaystyle c\, y = g\, \left(t\right), with nonzero constant a, is linear. It is homogeneous when g is identically zero.
Characteristic polynomial
For the homogeneous constant-coefficient equation, substituting exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) gives a r squared+b r+c\displaystyle b\, r + c. Its roots determine the homogeneous solution modes.
State system
Writing u=y\displaystyle u = y and v=y\displaystyle v = y prime turns one second-order equation into two first-order equations, so the state records both the value and its velocity.

Conditions before calculation

  • The elementary characteristic-polynomial method applies directly to homogeneous equations with constant coefficients, not arbitrary variable-coefficient equations.
  • Distinct real roots give two exponentials; a repeated root requires exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) and t exp⁡ (rt)\displaystyle t\, \operatorname{exp}\, \left(rt\right); conjugate roots alpha plus or minus i beta give real sine and cosine modes times exp⁡\displaystyle \operatorname{exp}(alpha t).
  • For the homogeneous constant-coefficient system x′\displaystyle x'=Ax, all eigenvalues with strictly negative real parts imply asymptotic stability of zero. Zero real parts require further analysis, including possible repeated-root growth.

Why the method works

1. Choose a complete family

A second-order homogeneous equation needs two independent solutions. Writing the same exponential twice provides only one adjustable direction and cannot satisfy arbitrary initial data.

2. Fit both initial conditions

Differentiate the proposed general solution before using the initial velocity. The resulting two linear equations determine its two constants.

3. Connect to a system

A forcing term enters the velocity equation. For a nonhomogeneous problem, add one particular solution to the full homogeneous family.

(uv)′=(01−c/a−b/a)(uv)+(0g(t)/a)\begin{pmatrix}u\\v\end{pmatrix}'=\begin{pmatrix}0&1\\-c/a&-b/a\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}+\begin{pmatrix}0\\g(t)/a\end{pmatrix}

Worked examples

EXAMPLE 1

Solve y double prime+3y prime+2 y=0\displaystyle 2\, y = 0, with y (0)=1\displaystyle y\, \left(0\right) = 1 and y prime(0)=0.

  1. Factor the characteristic polynomial to find two distinct decay rates.

    r2+3r+2=(r+1)(r+2)r^2+3r+2=(r+1)(r+2)

  2. Write the general solution and its initial-value equations.

    y=C1e−t+C2e−2t,C1+C2=1,−C1−2C2=0y=C_1e^{-t}+C_2e^{-2t},\quad C_1+C_2=1,\quad -C_1-2C_2=0

  3. Solving gives C 1=2\displaystyle C\, 1 = 2 and C 2=−1\displaystyle C\, 2 = -1. Each exponential satisfies the differential equation, and the two initial conditions check.

Conclusion: The solution is 2 exp⁡ (−t)−exp⁡ (−2 t)\displaystyle 2\, \operatorname{exp}\, \left(-t\right) - \operatorname{exp}\, \left(-2\, t\right), which tends to zero as t increases.

EXAMPLE 2

Solve y double prime-2y prime+y=0\displaystyle y = 0, with y (0)=0\displaystyle y\, \left(0\right) = 0 and y prime(0)=1.

  1. The characteristic polynomial has a repeated root.

    r2−2r+1=(r−1)2r^2-2r+1=(r-1)^2

  2. Use an independent second solution; y (0)=0\displaystyle y\, \left(0\right) = 0 forces C 1=0\displaystyle C\, 1 = 0.

    y=(C1+C2t)ety=(C_1+C_2t)e^t

  3. Differentiate to obtain y prime=C 2 (1+t) exp⁡ (t)\displaystyle C\, 2\, \left(1 + t\right)\, \operatorname{exp}\, \left(t\right); the initial derivative gives C 2=1\displaystyle C\, 2 = 1. Direct substitution yields zero.

    y=tet,y′′−2y′+y=[(t+2)−2(t+1)+t]et=0y=te^t,\quad y''-2y'+y=[(t+2)-2(t+1)+t]e^t=0

Conclusion: The solution grows without bound; the repeated positive root produces an unstable mode.

Pitfalls & counterexamples

Tempting claim: Purely imaginary roots make every linear system asymptotically stable.

Correction: They can produce persistent oscillation rather than decay; repeated structures can also cause growth.

Test the claim: The equation y double prime+y=0\displaystyle y = 0 has the nondecaying solution cos⁡ (t)\displaystyle \operatorname{cos}\, \left(t\right).

Check your understanding

What extra factor is needed for the second solution at a repeated real root r?

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Compare your reasoning

A factor of t.

The functions exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) and t exp⁡ (rt)\displaystyle t\, \operatorname{exp}\, \left(rt\right) are independent, unlike two constant multiples of exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right).

Connect this idea

Eigenvalues of the companion matrix equal the characteristic roots. Mechanical damping, circuits, and coupled linear systems use the same mode analysis; nonlinear stability needs additional hypotheses.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For the equation y″−5 y′+6 y=0\displaystyle -5\, y' + 6\, y = 0, find the larger characteristic root.

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  2. Problem 2

    For y″=0 with y (0)=1\displaystyle y\, \left(0\right) = 1 and y′ (0)=3\displaystyle y'\, \left(0\right) = 3, find y(2).

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  3. Problem 3

    A diagonal system has u′=−2 u\displaystyle u' = -2\, u and v′=1 v\displaystyle v' = 1\, v. Is its zero equilibrium asymptotically stable? Enter 1 for yes, 0 for no.

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  4. Problem 4

    The ODE y″+4 y=0\displaystyle +4\, y = 0. In its solution C1 cos⁡ (ω x)+C2 sin⁡ (ω x)\displaystyle C_{1}\, \operatorname{cos}\, \left(\omega\, x\right) + C_{2}\, \operatorname{sin}\, \left(\omega\, x\right), what is the positive angular frequency ω\displaystyle \omega?

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  5. Problem 5New context

    An exponential mode of the equation y″−7 y′+12 y=0\displaystyle -7\, y' + 12\, y = 0, find the larger characteristic root.

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  6. Problem 6New context

    A response satisfies y″=0 with y (0)=2\displaystyle y\, \left(0\right) = 2 and y′ (0)=3\displaystyle y'\, \left(0\right) = 3, find y(2).

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  7. Problem 7New context

    A two-component model has u′=−3 u\displaystyle u' = -3\, u and v′=2 v\displaystyle v' = 2\, v. Is its zero equilibrium asymptotically stable? Enter 1 for yes, 0 for no.

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  8. Problem 8New context

    An oscillator obeys y″+9 y=0\displaystyle +9\, y = 0. In its solution C1 cos⁡ (ω x)+C2 sin⁡ (ω x)\displaystyle C_{1}\, \operatorname{cos}\, \left(\omega\, x\right) + C_{2}\, \operatorname{sin}\, \left(\omega\, x\right), what is the positive angular frequency ω\displaystyle \omega?

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STAGE UODE 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Laplace Transforms & Euler Steps

Useful preparation: Second-Order Equations & Linear Systems

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A Laplace transform converts a time-dependent equation into an algebraic equation in a new variable s. Integration by parts turns a derivative into multiplication by s and an initial-value term. Euler stepping approaches the same initial-value problem locally: follow the current tangent for a short time, then recompute the slope at the new approximate point.

Unilateral Laplace transform
For a function f defined for nonnegative time, its transform F(s) is the integral of exp⁡ (−st) f (t)\displaystyle \operatorname{exp}\, \left(-st\right)\, f\, \left(t\right) from zero to infinity, whenever that improper integral converges.
Derivative rule
If Y is the transform of y and the relevant integrals exist, the transform of y prime is sY-y(0). Initial conditions therefore become part of the algebra.
Explicit Euler method
For y prime=f(t,y), step size h gives y at the next grid point approximately as the current y plus h times the current slope.

Conditions before calculation

  • Piecewise continuity on finite intervals and an exponential growth bound are sufficient for transform existence for sufficiently large real s. Transform existence is not automatic.
  • The derivative rule requires suitable regularity of y and convergence of the transforms, with the boundary term at infinity vanishing; use the right-hand initial value at zero.
  • Euler is a numerical approximation. Under standard smoothness and Lipschitz assumptions it has global error of order h over a fixed finite interval; stability can still force a small step.

Why the method works

1. Keep the initial term

Integration by parts explains the minus sign at zero. Dropping this term solves a different initial-value problem.

L{y′}=∫0∞e−sty′(t) dt=sY(s)−y(0)\mathcal L\{y'\}=\int_0^\infty e^{-st}y'(t)\,dt=sY(s)-y(0)

2. Solve and invert

After algebraically isolating Y, split it into known transforms. Check the recovered time function in both the original equation and the initial condition.

3. Assess numerical behavior

Euler uses the slope at the old state. For decay y prime=-k y with positive k, the multiplier must have magnitude below one to decay numerically.

yn+1=yn+hf(tn,yn),∣1−kh∣<1y_{n+1}=y_n+h f(t_n,y_n),\qquad |1-kh|<1

Worked examples

EXAMPLE 1

Use a Laplace transform to solve y prime+3 y=6\displaystyle 3\, y = 6 with y (0)=1\displaystyle y\, \left(0\right) = 1.

  1. Transform both sides, retaining the initial value and the transform of a constant.

    (sY−1)+3Y=6s(sY-1)+3Y=\frac6s

  2. Isolate Y and decompose it into two known transforms.

    Y=1+6/ss+3=2s−1s+3Y=\frac{1+6/s}{s+3}=\frac2s-\frac1{s+3}

  3. Invert and verify the differential equation and initial value.

    y(t)=2−e−3t,y′+3y=3e−3t+6−3e−3t=6,y(0)=1y(t)=2-e^{-3t},\qquad y'+3y=3e^{-3t}+6-3e^{-3t}=6,\quad y(0)=1

Conclusion: The solution starts at 1 and approaches equilibrium 2 for increasing nonnegative time.

EXAMPLE 2

Take two Euler steps for y prime=-3y, y (0)=2\displaystyle y\, \left(0\right) = 2, with h=0.1\displaystyle h = 0.1.

  1. The current-slope rule gives a multiplier of 0.7.

    yn+1=yn−0.3yn=0.7yny_{n+1}=y_n-0.3y_n=0.7y_n

  2. Apply the update twice, always using the previous approximate value.

    y1=1.4,y2=0.98,t2=0.2y_1=1.4,\qquad y_2=0.98,\qquad t_2=0.2

  3. Compare with the exact exponential at the same time.

    y(0.2)=2e−0.6≈1.097623,∣y2−y(0.2)∣≈0.117623y(0.2)=2e^{-0.6}\approx1.097623,\qquad |y_2-y(0.2)|\approx0.117623

Conclusion: Euler underestimates this decaying exponential here; the computed value is an approximation, even though the update arithmetic is exact.

Pitfalls & counterexamples

Tempting claim: A stable differential equation stays stable under any Euler step size.

Correction: The discretization has its own stability restriction.

Test the claim: For y prime=-3y, a step h=1\displaystyle h = 1 gives multiplier -2, so the numerical magnitudes double while the exact solution decays.

Check your understanding

What is the Laplace transform of y double prime when y (0)=a\displaystyle y\, \left(0\right) = a and y prime(0)=b?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

s squared Y minus sa minus b.

Apply the first-derivative rule twice, retaining both initial values.

Connect this idea

Transforms are especially useful for linear systems and switched forcing. Euler links derivatives to algorithms; better accuracy and stiffness handling motivate higher-order and implicit methods.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    For f (t)=e2 t\displaystyle f\, \left(t\right) = e^{2\, t}, evaluate its Laplace transform at s=5\displaystyle s = 5.

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  2. Problem 2

    For y′=0\displaystyle y' = 0 with y (0)=3\displaystyle y\, \left(0\right) = 3, the transformed equation is sY=b. Find b.

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  3. Problem 3

    Use one Euler step for y′=2 y\displaystyle y' = 2\, y, y (0)=2\displaystyle y\, \left(0\right) = 2, with step h=14\displaystyle h = \frac{1}{4}. What is the approximate y (14)\displaystyle y\, \left(\frac{1}{4}\right)?

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  4. Problem 4

    Use two Euler steps for y′=−y\displaystyle y' = -y, y (0)=8\displaystyle y\, \left(0\right) = 8, with h=12\displaystyle h = \frac{1}{2}. What is y2\displaystyle y_{2}?

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  5. Problem 5New context

    A transformed input is f (t)=e3 t\displaystyle f\, \left(t\right) = e^{3\, t}, evaluate its Laplace transform at s=6\displaystyle s = 6.

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  6. Problem 6New context

    A constant-state initial-value problem has y′=0\displaystyle y' = 0 with y (0)=4\displaystyle y\, \left(0\right) = 4, the transformed equation is sY=b. Find b.

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  7. Problem 7New context

    A numerical growth forecast uses one Euler step for y′=2 y\displaystyle y' = 2\, y, y (0)=3\displaystyle y\, \left(0\right) = 3, with step h=14\displaystyle h = \frac{1}{4}. What is the approximate y (14)\displaystyle y\, \left(\frac{1}{4}\right)?

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  8. Problem 8New context

    A numerical decay forecast uses two Euler steps for y′=−y\displaystyle y' = -y, y (0)=12\displaystyle y\, \left(0\right) = 12, with h=12\displaystyle h = \frac{1}{2}. What is y2\displaystyle y_{2}?

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