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.

Contents6 sections

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 $\dfrac{dy}{dx} = 3x^2 y$ .

Find the general solution by separating variables.
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 $\dfrac{dy}{dx} = \cos x \cdot e^{-y}$ .

Find the general implicit solution.
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 $y(0) = 1$ .

Separate variables and find the general solution.
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 $x > 1$ , $y > 1$ .

Use partial fractions on both sides to find the general solution in implicit form.
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 $x > 0$ .

Identify $P(x)$ and compute the integrating factor $\mu(x)$ .
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 $y^{\prime} - 3y = e^{5x}, \quad y(0) = 1$ .

Find the general solution using an integrating factor.
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.

Set up the differential equation for $Q(t)$ , the amount of salt (in kg) in the tank at time $t$ (in minutes).
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 $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^3$ .

Apply the substitution $v = y^{1-3} = y^{-2}$ to reduce the equation to a linear ODE in $v$ .
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 $y^{\prime\prime} - 5y^{\prime} + 6y = 0$ .

Write and solve the characteristic equation.
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 $y^{\prime\prime} + 4y^{\prime} + 4y = 0, \quad y(0) = 3, \quad y^{\prime}(0) = -1$ .

Find the repeated root and write the general solution.
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 $y^{\prime\prime} - 2y^{\prime} + 5y = 0, \quad y(0) = 0, \quad y^{\prime}(0) = 4$ .

Find the complex roots and write the general real-valued solution using sine and cosine.
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 $y_2 = e^{-x}\sin 3x$ .

Identify the complex roots from which these solutions arise and reconstruct the characteristic equation.
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 $y^{\prime\prime} + 3y^{\prime} + 2y = 4x + 1$ .

Solve the associated homogeneous equation.
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 $y^{\prime\prime} - 4y^{\prime} + 4y = 3e^{2x}$ .

Solve the homogeneous equation and explain why the naive trial $y_p = Ae^{2x}$ fails.
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 $y^{\prime\prime} + 9y = 2\sin 3x, \quad y(0) = 0, \quad y^{\prime}(0) = 0$ .

Note why a naive trial $y_p = A\cos 3x + B\sin 3x$ leads to resonance and state the correct trial function.
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 $y^{\prime\prime} - y = e^x$ .

Identify $y_1$ , $y_2$ , and compute the Wronskian.
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 $-\pi/2 < x < \pi/2$ .

State the homogeneous solution and set up the variation of parameters formulas.
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 $x > 0$ .

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.
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.

Write and solve the governing differential equation for $A(t)$ , the amount remaining at time $t$ .
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.

Set up and solve the IVP for the temperature $T(t)$ using Newton’s law of cooling.
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.

Write the IVP governing the displacement $x(t)$ and classify the damping type.
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.