Differential Equations: Problems & Exercises with solutions

This page collects carefully crafted differential equations problems organized by topic and difficulty. Whether you are studying separable equations, first-order linear ODEs, initial value problems, second-order homogeneous equations, characteristic equations, undetermined coefficients, or variation of parameters, each section below follows a pedagogical progression from easy to hard, with hints and full step-by-step solutions.

Separable Differential Equations

A differential equation is separable if it can be written in the form \( \frac{dy}{dx} = f(x)\,g(y) \). The technique requires isolating all \(y\)-terms on one side and all \(x\)-terms on the other, then integrating both sides. These problems form the foundation of first-order ODE methods and appear consistently on calculus and differential equations examinations.

Problem 1: Basic Separation of Variables

Easy

Consider the differential equation \( \dfrac{dy}{dx} = 3x^2 y \).

  1. Find the general solution by separating variables.
  2. Find the particular solution satisfying \( y(0) = 2 \).
Hint
Divide both sides by \(y\) (assuming \(y \neq 0\)) and multiply both sides by \(dx\). You will then be able to integrate \(\int \frac{1}{y}\,dy\) on the left and \(\int 3x^2\,dx\) on the right. For part 2, substitute the initial condition into the general solution to solve for the constant \(C\).
View Solution
Solution to question 1:

Separate variables:

\[ \frac{1}{y}\,dy = 3x^2\,dx \]

Integrate both sides:

\[ \ln|y| = x^3 + C_1 \]

Exponentiate both sides:

\[ y = Ce^{x^3}, \quad C \in \mathbb{R},\; C \neq 0 \]
Solution to question 2:

Apply \( y(0) = 2 \):

\[ 2 = Ce^{0} \implies C = 2 \]
\[ \boxed{y = 2e^{x^3}} \]

Problem 2: Separation with a Trigonometric Right-Hand Side

Easy

Consider the differential equation \( \dfrac{dy}{dx} = \cos x \cdot e^{-y} \).

  1. Find the general implicit solution.
  2. Solve explicitly for \(y\) and apply the initial condition \( y(0) = 0 \).
Hint
Multiply both sides by \(e^{y}\) to collect all \(y\)-terms on the left. The right side becomes \(\cos x\,dx\). Integrate both sides and then isolate \(y\) using the natural logarithm.
View Solution
Solution to question 1:

Rewrite as \( e^{y}\,dy = \cos x\,dx \) and integrate:

\[ e^{y} = \sin x + C \]
Solution to question 2:

Take the natural log to solve explicitly:

\[ y = \ln(\sin x + C) \]

Apply \( y(0) = 0 \): \(\ln(0 + C) = 0 \implies C = 1\).

\[ \boxed{y = \ln(\sin x + 1)} \]

Problem 3: Rational Separable Equation with an Initial Value

Medium

Consider the differential equation \( (1 + x^2)\,\dfrac{dy}{dx} = x\,y^2 \) with initial condition \( y(0) = 1 \).

  1. Separate variables and find the general solution.
  2. Apply the initial condition and determine the interval of validity of the solution.
Hint
Divide both sides by \(y^2(1+x^2)\). On the \(x\)-side you will need to integrate \(\int \frac{x}{1+x^2}\,dx\); consider the substitution \(u = 1 + x^2\). To find the interval of validity, determine where the particular solution is defined and continuous.
View Solution
Solution to question 1:

Separate:

\[ \frac{dy}{y^2} = \frac{x\,dx}{1+x^2} \]

Integrate both sides:

\[ -\frac{1}{y} = \frac{1}{2}\ln(1+x^2) + C \]

General solution (implicit):

\[ y = \frac{-1}{\tfrac{1}{2}\ln(1+x^2) + C} \]
Solution to question 2:

Apply \(y(0) = 1\): \(-1 = C\), so \(C = -1\).

\[ y = \frac{1}{1 – \tfrac{1}{2}\ln(1+x^2)} \]

The solution is valid as long as the denominator is nonzero: \(\ln(1+x^2) \neq 2\), i.e., \(x^2 \neq e^2 – 1\). The interval of validity containing \(x = 0\) is:

\[ \boxed{-\sqrt{e^2 – 1} < x < \sqrt{e^2-1}} \]

Problem 4: Bernoulli-Type Separable Equation

Hard

Consider the equation \( \dfrac{dy}{dx} = \dfrac{y^2 – 1}{x^2 – 1} \) for \(x > 1\), \(y > 1\).

  1. Use partial fractions on both sides to find the general solution in implicit form.
  2. Verify that \(y = x\) is a particular solution and comment on how it fits into the general family.
Hint
After separating, you will integrate \(\int \frac{dy}{y^2-1}\) and \(\int \frac{dx}{x^2-1}\). Both require the same partial fraction decomposition: \(\frac{1}{t^2-1} = \frac{1}{2}\left(\frac{1}{t-1} – \frac{1}{t+1}\right)\). Simplify using logarithm rules after integrating.
View Solution
Solution to question 1:

Separate: \(\dfrac{dy}{y^2-1} = \dfrac{dx}{x^2-1}\). Apply partial fractions to each side:

\[ \frac{1}{2}\ln\left|\frac{y-1}{y+1}\right| = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C_1 \]

Multiply by 2 and exponentiate:

\[ \frac{y-1}{y+1} = K\,\frac{x-1}{x+1}, \quad K > 0 \]
Solution to question 2:

If \(y = x\), then \(\dfrac{y-1}{y+1} = \dfrac{x-1}{x+1}\), so \(K = 1\). This is indeed a member of the general family. Direct substitution confirms \(\dfrac{dy}{dx} = 1 = \dfrac{x^2-1}{x^2-1} = 1\). \(\checkmark\)

First-Order Linear Differential Equations and Integrating Factors

A first-order linear ODE has the standard form \( \dfrac{dy}{dx} + P(x)\,y = Q(x) \). The key technique is to multiply through by the integrating factor \( \mu(x) = e^{\int P(x)\,dx} \), which transforms the left side into an exact derivative. These equations model phenomena such as mixing problems, Newton’s law of cooling, and population growth with harvesting.

Problem 5: Standard Integrating Factor

Easy

Solve the differential equation \( \dfrac{dy}{dx} + \dfrac{2}{x}\,y = x^3 \) for \(x > 0\).

  1. Identify \(P(x)\) and compute the integrating factor \(\mu(x)\).
  2. Find the general solution.
Hint
With \(P(x) = 2/x\), the integrating factor is \(\mu = e^{\int (2/x)\,dx} = x^2\). Multiply the entire equation by \(x^2\) and recognize the left side as \(\dfrac{d}{dx}[x^2 y]\).
View Solution
Solution to question 1:

\(P(x) = \dfrac{2}{x}\), so \(\mu(x) = e^{2\ln x} = x^2\).

Solution to question 2:

Multiply through:

\[ \frac{d}{dx}\!\left[x^2 y\right] = x^5 \]

Integrate both sides:

\[ x^2 y = \frac{x^6}{6} + C \]
\[ \boxed{y = \frac{x^4}{6} + \frac{C}{x^2}} \]

Problem 6: Initial Value Problem with Exponential Forcing

Easy

Solve the initial value problem \( y^{\prime} – 3y = e^{5x}, \quad y(0) = 1 \).

  1. Find the general solution using an integrating factor.
  2. Apply the initial condition to find the particular solution.
Hint
Here \(P(x) = -3\), so \(\mu = e^{-3x}\). After multiplying through, the right side becomes \(e^{5x} \cdot e^{-3x} = e^{2x}\), which integrates easily.
View Solution
Solution to question 1:

Multiply by \(e^{-3x}\):

\[ \frac{d}{dx}\!\left[e^{-3x}y\right] = e^{2x} \]

Integrate:

\[ e^{-3x}y = \frac{e^{2x}}{2} + C \implies y = \frac{e^{5x}}{2} + Ce^{3x} \]
Solution to question 2:

At \(x=0\): \(1 = \tfrac{1}{2} + C \implies C = \tfrac{1}{2}\).

\[ \boxed{y = \frac{e^{5x} + e^{3x}}{2}} \]

Problem 7: Mixing Tank Problem

Medium

A tank initially contains 100 L of pure water. A salt solution with concentration 0.5 kg/L flows in at 4 L/min, and the well-mixed solution drains out at the same rate.

  1. Set up the differential equation for \(Q(t)\), the amount of salt (in kg) in the tank at time \(t\) (in minutes).
  2. Solve the IVP and find the limiting amount of salt in the tank as \(t \to \infty\).
Hint
Rate in = (concentration in) × (flow rate in). Rate out = (concentration out) × (flow rate out), where concentration out = \(Q(t)/100\). Write \(Q^{\prime} = \text{rate in} – \text{rate out}\) and bring it to standard linear form.
View Solution
Solution to question 1:
\[ Q^{\prime} = (0.5)(4) – \frac{Q}{100}(4) = 2 – \frac{Q}{25} \]

Standard form: \( Q^{\prime} + \dfrac{Q}{25} = 2 \), \(Q(0) = 0\).

Solution to question 2:

Integrating factor \(\mu = e^{t/25}\):

\[ \frac{d}{dt}\!\left[e^{t/25}Q\right] = 2e^{t/25} \]
\[ Q = 50 + Ce^{-t/25} \]

Apply \(Q(0)=0\): \(C = -50\).

\[ \boxed{Q(t) = 50\!\left(1 – e^{-t/25}\right)} \]

As \(t \to \infty\), \(Q \to 50\) kg, which equals the steady-state: 100 L \(\times\) 0.5 kg/L.

Problem 8: Bernoulli Equation

Hard

Solve the Bernoulli equation \( \dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^3 \).

  1. Apply the substitution \( v = y^{1-3} = y^{-2} \) to reduce the equation to a linear ODE in \(v\).
  2. Solve for \(v(x)\), then recover \(y(x)\).
Hint
Divide the equation by \(y^3\) first, so you have \(y^{-3}y^{\prime} + \frac{y^{-2}}{x} = x^2\). Then set \(v = y^{-2}\) and compute \(v^{\prime} = -2y^{-3}y^{\prime}\). Substituting converts the equation into a first-order linear ODE in \(v\) that you can solve with an integrating factor.
View Solution
Solution to question 1:

Divide by \(y^3\):

\[ y^{-3}y^{\prime} + \frac{y^{-2}}{x} = x^2 \]

Let \(v = y^{-2}\), so \(v^{\prime} = -2y^{-3}y^{\prime}\), giving \(y^{-3}y^{\prime} = -v^{\prime}/2\):

\[ -\frac{v^{\prime}}{2} + \frac{v}{x} = x^2 \implies v^{\prime} – \frac{2v}{x} = -2x^2 \]
Solution to question 2:

Integrating factor: \(\mu = x^{-2}\). Then:

\[ \frac{d}{dx}\!\left[\frac{v}{x^2}\right] = -2 \implies \frac{v}{x^2} = -2x + C \]
\[ v = -2x^3 + Cx^2 \]

Recover \(y\) from \(v = y^{-2}\):

\[ \boxed{y = \frac{1}{\sqrt{Cx^2 – 2x^3}}} \]

Second-Order Homogeneous Linear Differential Equations

A second-order homogeneous linear ODE with constant coefficients has the form \( ay^{\prime\prime} + by^{\prime} + cy = 0 \). Its general solution is built from the roots of the characteristic equation \( ar^2 + br + c = 0 \). The nature of those roots — distinct real, repeated real, or complex conjugate — determines the form of the general solution. Mastering these three cases is essential before moving to nonhomogeneous equations.

Problem 9: Distinct Real Roots

Easy

Solve the differential equation \( y^{\prime\prime} – 5y^{\prime} + 6y = 0 \).

  1. Write and solve the characteristic equation.
  2. State the general solution and find the particular solution with \( y(0) = 2 \), \( y^{\prime}(0) = 5 \).
Hint
Substitute \(y = e^{rx}\) and factor the resulting quadratic. With two distinct real roots \(r_1\) and \(r_2\), the general solution is \(y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\). Differentiate once to use both initial conditions.
View Solution
Solution to question 1:

Characteristic equation: \(r^2 – 5r + 6 = (r-2)(r-3) = 0\), so \(r_1 = 2,\; r_2 = 3\).

Solution to question 2:
\[ y = c_1 e^{2x} + c_2 e^{3x}, \quad y^{\prime} = 2c_1 e^{2x} + 3c_2 e^{3x} \]

At \(x=0\): \(c_1 + c_2 = 2\) and \(2c_1 + 3c_2 = 5\). Solving: \(c_1 = 1,\; c_2 = 1\).

\[ \boxed{y = e^{2x} + e^{3x}} \]

Problem 10: Repeated Root Case

Easy

Solve the IVP \( y^{\prime\prime} + 4y^{\prime} + 4y = 0, \quad y(0) = 3, \quad y^{\prime}(0) = -1 \).

  1. Find the repeated root and write the general solution.
  2. Apply the initial conditions.
Hint
The characteristic equation factors as a perfect square. When the characteristic equation yields a repeated root \(r = \alpha\), the two linearly independent solutions are \(e^{\alpha x}\) and \(x e^{\alpha x}\).
View Solution
Solution to question 1:

Characteristic equation: \(r^2 + 4r + 4 = (r+2)^2 = 0\), so \(r = -2\) (repeated).

\[ y = (c_1 + c_2 x)e^{-2x} \]
Solution to question 2:

Differentiate: \(y^{\prime} = c_2 e^{-2x} – 2(c_1 + c_2 x)e^{-2x}\).

At \(x = 0\): \(c_1 = 3\) and \(c_2 – 2c_1 = -1 \implies c_2 = 5\).

\[ \boxed{y = (3 + 5x)e^{-2x}} \]

Problem 11: Complex Conjugate Roots

Medium

Solve the IVP \( y^{\prime\prime} – 2y^{\prime} + 5y = 0, \quad y(0) = 0, \quad y^{\prime}(0) = 4 \).

  1. Find the complex roots and write the general real-valued solution using sine and cosine.
  2. Apply the initial conditions to determine the particular solution.
Hint
The discriminant is negative, so the roots are complex: \(r = \alpha \pm \beta i\). Use Euler’s formula to write the general solution as \(y = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)\).
View Solution
Solution to question 1:

Characteristic equation: \(r^2 – 2r + 5 = 0\). Discriminant: \(4 – 20 = -16\).

\[ r = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
\[ y = e^{x}(c_1 \cos 2x + c_2 \sin 2x) \]
Solution to question 2:

At \(x = 0\): \(y(0) = c_1 = 0\).

Differentiate with \(c_1 = 0\): \(y^{\prime} = e^x(c_2 \sin 2x) \cdot 1 + e^x(2c_2\cos 2x)\). At \(x=0\): \(y^{\prime}(0) = 2c_2 = 4 \implies c_2 = 2\).

\[ \boxed{y = 2e^{x}\sin 2x} \]

Problem 12: Constructing an Equation from Known Solutions

Medium

Two functions are given: \( y_1 = e^{-x}\cos 3x \) and \( y_2 = e^{-x}\sin 3x \).

  1. Identify the complex roots from which these solutions arise and reconstruct the characteristic equation.
  2. Write the second-order homogeneous ODE with constant coefficients having these as a fundamental set of solutions, and compute the Wronskian \(W(y_1, y_2)\).
Hint
From \(e^{\alpha x}\cos\beta x\) and \(e^{\alpha x}\sin\beta x\), read off \(\alpha = -1\) and \(\beta = 3\). The roots are \(r = -1 \pm 3i\). Reconstruct the characteristic polynomial as \((r – (-1+3i))(r – (-1-3i))\). For the Wronskian, use \(W = y_1 y_2^{\prime} – y_2 y_1^{\prime}\).
View Solution
Solution to question 1:

Roots: \(r = -1 \pm 3i\). Characteristic polynomial:

\[ (r+1)^2 + 9 = r^2 + 2r + 10 = 0 \]
Solution to question 2:

The ODE is \( y^{\prime\prime} + 2y^{\prime} + 10y = 0 \).

Wronskian:

\[ W = y_1 y_2^{\prime} – y_2 y_1^{\prime} \]

By Abel’s theorem, \(W(x) = W(0)\,e^{-\int_0^x 2\,dt} = W(0)\,e^{-2x}\). Computing \(W(0) = (1)(3) – (0)(-1) = 3\):

\[ \boxed{W(y_1, y_2) = 3e^{-2x} \neq 0} \]

Since \(W \neq 0\), the set \(\{y_1, y_2\}\) is indeed a fundamental set of solutions.

Nonhomogeneous Equations: Method of Undetermined Coefficients

When the right-hand side of \( ay^{\prime\prime} + by^{\prime} + cy = f(x) \) consists of polynomials, exponentials, sines, cosines, or their products, the method of undetermined coefficients offers an efficient path to finding a particular solution \(y_p\). The general solution is then \(y = y_h + y_p\), where \(y_h\) is the homogeneous solution. The key step is recognizing when the trial function overlaps with part of \(y_h\), requiring multiplication by \(x\) to avoid redundancy.

Problem 13: Polynomial Forcing Function

Easy

Find the general solution of \( y^{\prime\prime} + 3y^{\prime} + 2y = 4x + 1 \).

  1. Solve the associated homogeneous equation.
  2. Guess a particular solution of the form \( y_p = Ax + B \) and determine \(A\) and \(B\).
Hint
For a polynomial forcing term of degree 1, guess \(y_p = Ax + B\). Substitute \(y_p\), \(y_p^{\prime}\), and \(y_p^{\prime\prime}\) into the ODE and match coefficients of equal powers of \(x\) on both sides.
View Solution
Solution to question 1:

Characteristic equation: \(r^2 + 3r + 2 = (r+1)(r+2) = 0\), so \(r = -1, -2\).

\[ y_h = c_1 e^{-x} + c_2 e^{-2x} \]
Solution to question 2:

With \(y_p = Ax + B\), we have \(y_p^{\prime} = A\), \(y_p^{\prime\prime} = 0\). Substituting:

\[ 0 + 3A + 2(Ax + B) = 4x + 1 \implies 2Ax + (3A + 2B) = 4x + 1 \]

Matching: \(A = 2\) and \(3(2) + 2B = 1 \implies B = -\tfrac{5}{2}\).

\[ \boxed{y = c_1 e^{-x} + c_2 e^{-2x} + 2x – \frac{5}{2}} \]

Problem 14: Resonance — Repeated Trial Function

Medium

Find the general solution of \( y^{\prime\prime} – 4y^{\prime} + 4y = 3e^{2x} \).

  1. Solve the homogeneous equation and explain why the naive trial \( y_p = Ae^{2x} \) fails.
  2. Use the modified trial \( y_p = Ax^2 e^{2x} \) to find the particular solution.
Hint
The homogeneous solution contains \(e^{2x}\) and \(xe^{2x}\) (repeated root \(r=2\)). Because both overlap with \(Ae^{2x}\) and \(Axe^{2x}\), the trial function must be multiplied by \(x^2\) to avoid redundancy. Compute \(y_p^{\prime}\) and \(y_p^{\prime\prime}\) carefully using the product rule.
View Solution
Solution to question 1:

Characteristic equation: \((r-2)^2 = 0\), so \(y_h = (c_1 + c_2 x)e^{2x}\). Both \(e^{2x}\) and \(xe^{2x}\) are already in \(y_h\), so \(Ae^{2x}\) and \(Axe^{2x}\) yield zero when substituted.

Solution to question 2:

Let \(y_p = Ax^2 e^{2x}\). Then:

\[ y_p^{\prime} = A(2x + 2x^2)e^{2x}, \quad y_p^{\prime\prime} = A(2 + 8x + 4x^2)e^{2x} \]

Substitute into \(y^{\prime\prime} – 4y^{\prime} + 4y\):

\[ A(2 + 8x + 4x^2)e^{2x} – 4A(2x + 2x^2)e^{2x} + 4Ax^2 e^{2x} = 2Ae^{2x} = 3e^{2x} \]

So \(A = \tfrac{3}{2}\).

\[ \boxed{y = (c_1 + c_2 x)e^{2x} + \frac{3}{2}x^2 e^{2x}} \]

Problem 15: Sinusoidal Forcing and IVP

Hard

Solve the IVP \( y^{\prime\prime} + 9y = 2\sin 3x, \quad y(0) = 0, \quad y^{\prime}(0) = 0 \).

  1. Note why a naive trial \( y_p = A\cos 3x + B\sin 3x \) leads to resonance and state the correct trial function.
  2. Find the complete particular and general solution, then apply the initial conditions.
Hint
The homogeneous solution is \(c_1\cos 3x + c_2\sin 3x\), so the forcing frequency matches the natural frequency. Multiply the trial by \(x\): guess \(y_p = x(A\cos 3x + B\sin 3x)\). After computing derivatives, collect like terms and match coefficients.
View Solution
Solution to question 1:

The homogeneous solution is \(y_h = c_1\cos 3x + c_2\sin 3x\). The forcing term \(\sin 3x\) duplicates part of \(y_h\), so we use \(y_p = x(A\cos 3x + B\sin 3x)\).

Solution to question 2:

Computing \(y_p^{\prime\prime}\) and substituting into \(y^{\prime\prime} + 9y\) yields \(-6A\sin 3x + 6B\cos 3x = 2\sin 3x\). Matching:

\[ -6A = 2 \implies A = -\frac{1}{3}, \quad 6B = 0 \implies B = 0 \]
\[ y_p = -\frac{x}{3}\cos 3x \]

General solution: \(y = c_1\cos 3x + c_2\sin 3x – \dfrac{x}{3}\cos 3x\).

Apply \(y(0) = 0\): \(c_1 = 0\). Differentiate and apply \(y^{\prime}(0) = 0\): \(3c_2 – \tfrac{1}{3} = 0 \implies c_2 = \tfrac{1}{9}\).

\[ \boxed{y = \frac{\sin 3x}{9} – \frac{x\cos 3x}{3}} \]

Variation of Parameters

The method of variation of parameters finds a particular solution \(y_p = u_1(x)\,y_1 + u_2(x)\,y_2\) for any nonhomogeneous linear ODE \(y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = f(x)\), provided a fundamental set of solutions \(\{y_1, y_2\}\) of the homogeneous equation is known. Unlike undetermined coefficients, this method imposes no restrictions on the form of \(f(x)\). The formulas are \( u_1^{\prime} = -y_2 f / W \) and \( u_2^{\prime} = y_1 f / W \), where \(W\) is the Wronskian.

Problem 16: Variation of Parameters — Exponential Forcing

Medium

Use variation of parameters to find a particular solution of \( y^{\prime\prime} – y = e^x \).

  1. Identify \(y_1\), \(y_2\), and compute the Wronskian.
  2. Find \(u_1\) and \(u_2\), then write \(y_p\).
Hint
The homogeneous equation \(y^{\prime\prime} – y = 0\) has characteristic roots \(r = \pm 1\), giving \(y_1 = e^x\) and \(y_2 = e^{-x}\). Here \(f(x) = e^x\). One of the integrals \(\int u_1^{\prime}\,dx\) will need to be handled carefully because \(y_1 = e^x\) and \(f(x) = e^x\) have the same exponential.
View Solution
Solution to question 1:
\[ W = e^x(-e^{-x}) – e^{-x}(e^x) = -1 – 1 = -2 \]
Solution to question 2:
\[ u_1^{\prime} = -\frac{y_2 f}{W} = -\frac{e^{-x}\cdot e^x}{-2} = \frac{1}{2} \implies u_1 = \frac{x}{2} \]
\[ u_2^{\prime} = \frac{y_1 f}{W} = \frac{e^x \cdot e^x}{-2} = -\frac{e^{2x}}{2} \implies u_2 = -\frac{e^{2x}}{4} \]
\[ y_p = \frac{x}{2}e^x – \frac{e^{2x}}{4}e^{-x} = \frac{x e^x}{2} – \frac{e^x}{4} \]

The term \(-\tfrac{e^x}{4}\) is part of \(y_h\), so the essential particular solution is:

\[ \boxed{y_p = \frac{x e^x}{2}} \]

Problem 17: Variation of Parameters — Trigonometric Forcing

Medium

Find the general solution of \( y^{\prime\prime} + y = \sec x \) on \( -\pi/2 < x < \pi/2 \).

  1. State the homogeneous solution and set up the variation of parameters formulas.
  2. Evaluate the necessary integrals and write the complete general solution.
Hint
Here \(y_1 = \cos x\), \(y_2 = \sin x\), and \(W = 1\). The forcing function \(f = \sec x\) cannot be handled by undetermined coefficients, which makes this a natural candidate for variation of parameters. You will need \(\int \sin x \sec x\,dx\) and \(\int \cos x \sec x\,dx\).
View Solution
Solution to question 1:

\(y_h = c_1\cos x + c_2\sin x\), \(W = \cos^2 x + \sin^2 x = 1\).

\[ u_1^{\prime} = -\frac{\sin x \cdot \sec x}{1} = -\tan x, \quad u_2^{\prime} = \frac{\cos x \cdot \sec x}{1} = 1 \]
Solution to question 2:
\[ u_1 = \ln|\cos x|, \quad u_2 = x \]
\[ y_p = \cos x \ln(\cos x) + x\sin x \]
\[ \boxed{y = c_1\cos x + c_2\sin x + \cos x\ln(\cos x) + x\sin x} \]

Problem 18: Variation of Parameters with Non-Constant Coefficients (Euler Equation)

Hard

Consider the Euler (Cauchy–Euler) equation \( x^2 y^{\prime\prime} – 2xy^{\prime} + 2y = x^3\ln x \) for \(x > 0\).

  1. Verify that \(y_1 = x\) and \(y_2 = x^2\) are solutions of the associated homogeneous equation \(x^2 y^{\prime\prime} – 2xy^{\prime} + 2y = 0\) and compute their Wronskian.
  2. Rewrite the equation in standard form and apply variation of parameters to find the general solution.
Hint
To bring the equation to standard form \(y^{\prime\prime} + Py^{\prime} + Qy = f\), divide through by \(x^2\). The forcing term in standard form is \(f(x) = x\ln x\). With \(W = x^2\), compute \(u_1^{\prime} = -y_2 f/W\) and \(u_2^{\prime} = y_1 f/W\), then integrate by parts.
View Solution
Solution to question 1:

For \(y_1 = x\): \(y_1^{\prime\prime} = 0\), so \(x^2(0) – 2x(1) + 2x = 0\). \(\checkmark\)

For \(y_2 = x^2\): \(y_2^{\prime\prime} = 2\), so \(x^2(2) – 2x(2x) + 2x^2 = 2x^2 – 4x^2 + 2x^2 = 0\). \(\checkmark\)

\[ W = x(2x) – x^2(1) = 2x^2 – x^2 = x^2 \]
Solution to question 2:

Standard form: \(y^{\prime\prime} – \tfrac{2}{x}y^{\prime} + \tfrac{2}{x^2}y = x\ln x\).

\[ u_1^{\prime} = -\frac{x^2 \cdot x\ln x}{x^2} = -x\ln x \implies u_1 = -\frac{x^2}{2}\ln x + \frac{x^2}{4} \]
\[ u_2^{\prime} = \frac{x \cdot x\ln x}{x^2} = \ln x \implies u_2 = x\ln x – x \]
\[ y_p = u_1 x + u_2 x^2 = \left(-\frac{x^2\ln x}{2} + \frac{x^2}{4}\right)x + (x\ln x – x)x^2 \]
\[ y_p = \frac{x^3}{2}\ln x – \frac{3x^3}{4} \]
\[ \boxed{y = c_1 x + c_2 x^2 + \frac{x^3}{2}\ln x – \frac{3x^3}{4}} \]

Applications of Differential Equations

Differential equations model a wide range of real-world phenomena. This section presents applied problems drawn from exponential growth and decay, Newton’s law of cooling, and mechanical vibrations (spring-mass systems). Setting up the governing ODE from a physical description is itself a critical skill, tested frequently in engineering and applied mathematics courses.

Problem 19: Radioactive Decay and Half-Life

Easy

A radioactive substance decays at a rate proportional to the amount present. The half-life of the substance is 8 years.

  1. Write and solve the governing differential equation for \(A(t)\), the amount remaining at time \(t\).
  2. If the initial amount is 200 g, how much remains after 20 years? Give an exact answer and a decimal approximation.
Hint
The governing equation is \(A^{\prime} = kA\) with \(k < 0\). Use the half-life condition \(A(8) = A_0/2\) to determine \(k = -\ln 2 / 8\).
View Solution
Solution to question 1:
\[ A^{\prime} = kA \implies A(t) = A_0 e^{kt} \]

From \(A(8) = A_0/2\): \(e^{8k} = 1/2 \implies k = -\dfrac{\ln 2}{8}\).

\[ A(t) = A_0\, e^{-(\ln 2\,/\,8)\,t} = A_0 \cdot 2^{-t/8} \]
Solution to question 2:
\[ A(20) = 200 \cdot 2^{-20/8} = 200 \cdot 2^{-5/2} = \frac{200}{4\sqrt{2}} = \frac{50}{\sqrt{2}} = 25\sqrt{2} \approx 35.36 \text{ g} \]
\[ \boxed{A(20) = 25\sqrt{2} \approx 35.4 \text{ g}} \]

Problem 20: Newton’s Law of Cooling

Medium

A cup of coffee at 90°C is placed in a room maintained at 20°C. After 10 minutes, the coffee has cooled to 70°C.

  1. Set up and solve the IVP for the temperature \(T(t)\) using Newton’s law of cooling.
  2. Find the time at which the coffee reaches 40°C.
Hint
Newton’s law of cooling: \(\dfrac{dT}{dt} = k(T – T_{\text{room}})\). Let \(u = T – 20\) to get a simpler separable equation. After solving, use \(T(10) = 70\) to find \(k\), then solve \(T(t^*) = 40\) for \(t^*\).
View Solution
Solution to question 1:

Let \(u = T – 20\). Then \(u^{\prime} = ku\), so \(u = u_0 e^{kt}\). At \(t = 0\): \(u_0 = 70\). Thus \(T = 20 + 70e^{kt}\).

From \(T(10) = 70\): \(70e^{10k} = 50 \implies k = \dfrac{\ln(5/7)}{10}\).

\[ T(t) = 20 + 70\left(\frac{5}{7}\right)^{t/10} \]
Solution to question 2:

Set \(T = 40\): \(70\left(\tfrac{5}{7}\right)^{t/10} = 20 \implies \left(\tfrac{5}{7}\right)^{t/10} = \tfrac{2}{7}\).

\[ \frac{t}{10} = \frac{\ln(2/7)}{\ln(5/7)} \implies t = \frac{10\ln(2/7)}{\ln(5/7)} \approx 38.1 \text{ min} \]
\[ \boxed{t \approx 38.1 \text{ minutes}} \]

Problem 21: Damped Spring-Mass System

Hard

A mass of 1 kg is attached to a spring with spring constant \(k = 13\) N/m. The system experiences damping with coefficient \(c = 4\) N·s/m. At \(t = 0\) the mass is displaced 0.5 m from equilibrium and released from rest.

  1. Write the IVP governing the displacement \(x(t)\) and classify the damping type.
  2. Solve the IVP completely and describe the long-term behavior of the solution.
Hint
Newton’s second law gives \(mx^{\prime\prime} + cx^{\prime} + kx = 0\). With \(m=1\), \(c=4\), \(k=13\), compute the discriminant \(c^2 – 4mk\) to classify damping. A negative discriminant indicates underdamping, leading to complex roots and oscillatory decay.
View Solution
Solution to question 1:

The IVP is \( x^{\prime\prime} + 4x^{\prime} + 13x = 0, \quad x(0) = 0.5, \quad x^{\prime}(0) = 0 \).

Discriminant: \(4^2 – 4(1)(13) = 16 – 52 = -36 < 0\). The system is underdamped.

Solution to question 2:

Characteristic roots: \(r = \dfrac{-4 \pm 6i}{2} = -2 \pm 3i\).

\[ x(t) = e^{-2t}(c_1\cos 3t + c_2\sin 3t) \]

Apply \(x(0) = 0.5\): \(c_1 = 0.5\).

Apply \(x^{\prime}(0) = 0\): \(x^{\prime}(0) = -2c_1 + 3c_2 = 0 \implies c_2 = \tfrac{1}{3}\).

\[ \boxed{x(t) = e^{-2t}\!\left(\frac{1}{2}\cos 3t + \frac{1}{3}\sin 3t\right)} \]

As \(t \to \infty\), the factor \(e^{-2t} \to 0\), so the displacement decays to zero. The solution oscillates with pseudo-period \(\tfrac{2\pi}{3}\) seconds while the amplitude shrinks exponentially — the hallmark of underdamped oscillation.