Differential Equations

Mathematics · FE Reference Handbook section

Mathematics

10 formulas

10 exam-style examples

~60 min

All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A common class of ordinary linear differential equations is
where rn is the nth distinct root of the characteristic polynomial P(x) with
Higher orders of multiplicity imply higher powers of x. The complete solution for the differential equation is
where yp(x) is any particular solution with f(x) present. If f(x) has ern x terms, then resonance is manifested.
Furthermore, specific f(x) forms result in specific yp(x) forms, some of which are:
If the independent variable is time t, then transient dynamic solutions are implied.
First-Order Linear Homogeneous Differential Equations with Constant Coefficients

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Exponential decay — solve for value at t — Differential Equations

A mathematics problem uses Exponential decay. Given initial value (y0) = 19.5000; decay constant (k) = 1.6000 1/s; time (t) = 3.0000 s, determine the value at t (y).

Given

$initial value (y_{0}) = 19.5000$
$decay constant (k) = 1.6000 1/s$
$time (t) = 3.0000 s$

Find

value at t (y)

Start with the thinking

The governing relation printed in this handbook section is Exponential decay.
Everything except y is given, so isolate y symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange the relation so that y stands alone on the left-hand side.
Step 3 — List the givens: initial value (y0) = 19.5000, decay constant (k) = 1.6000 1/s, time (t) = 3.0000 s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$y = 0.1605$
Step 6 — Check: returning y = 0.1605 to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = 0.1605

Why the other options are there

0.3210 — kept a factor of two that cancels in the correct rearrangement.
0.0802 — dropped that same factor in the other direction.
0.1765 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Differential Equations

Example 2

First-order linear ODE solution — solve for value at time t — Differential Equations (2)

radioactive tracer decay in a groundwater test Given initial value (y_0) = 8.0000 mg/L; rate constant (k) = 0.4000 1/h; elapsed time (t) = 3.5000 h, determine the value at time t (y) in mg/L.

Given

$initial value (y_0) = 8.0000 mg/L$
$rate constant (k) = 0.4000 1/h$
$elapsed time (t) = 3.5000 h$

Find

value at time t (y), in mg/L

Start with the thinking

The governing relation printed in this handbook section is First-order linear ODE solution.
Everything except y is given, so isolate y symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The ODE y' + k y = 0 has the exponential solution shown.

Step-by-step solution

Step 1 — State the governing relation:
$y(t) = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for y:
$y = y = y_0 e^{-k t}$
Step 3 — List the givens: initial value (y_0) = 8.0000 mg/L, rate constant (k) = 0.4000 1/h, elapsed time (t) = 3.5000 h.
Step 4 — Substitute the given values:
$y = y = y_0 e^{-0.4000 3.5000}$
Step 5 — Evaluate:
$y = 1.9728\ \text{mg/L}$
Step 6 — Check: returning y = 1.9728 mg/L to
$y(t) = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = 1.9728\ \text{mg/L}

Why the other options are there

3.9456 — kept a factor of two that cancels in the correct rearrangement.
0.9864 — dropped that same factor in the other direction.
2.1701 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations (First Order)

Example 3

First-order linear differential equation solution — solve for solution value — Differential Equations (3)

The differential equation for radioactive decay is solved for the initial quantity. Given initial condition (y0) = 44.5000; rate constant (k) = 1.3500 1/s; time (t) = 0.8000 s, determine the solution value (y).

Given

$initial condition (y_{0}) = 44.5000$
$rate constant (k) = 1.3500 1/s$
$time (t) = 0.8000 s$

Find

solution value (y)

Start with the thinking

The governing relation printed in this handbook section is First-order linear differential equation solution.
Everything except y is given, so isolate y symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
This first-order differential equation dy/dt = -k y has the exponential decay solution shown.

Figure 3 — schematic for First-order linear differential equation solution — solve for solution value — Differential Equations (3)

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for y:
$y = y_0 e^{-kt}$
Step 3 — List the givens: initial condition (y0) = 44.5000, rate constant (k) = 1.3500 1/s, time (t) = 0.8000 s.
Step 4 — Substitute the given values:
$y = y_0 e^{-k0.8000}$
Step 5 — Evaluate:
$y = 15.1120$
Step 6 — Check: returning y = 15.1120 to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y = 15.1120

Why the other options are there

30.2240 — kept a factor of two that cancels in the correct rearrangement.
7.5560 — dropped that same factor in the other direction.
16.6232 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations

Example 4

Exponential decay — solve for initial value — Differential Equations (4)

A mathematics problem uses Exponential decay. Given decay constant (k) = 0.3000 1/s; time (t) = 1.3500 s; value at t (y) = 13.4600, determine the initial value (y0).

Given

$decay constant (k) = 0.3000 1/s$
$time (t) = 1.3500 s$
$value at t (y) = 13.4600$

Find

initial value (y0)

Start with the thinking

The governing relation printed in this handbook section is Exponential decay.
Everything except y0 is given, so isolate y0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange the relation so that y0 stands alone on the left-hand side.
Step 3
$List the givens: decay constant (k) = 0.3000 1/s, time (t) = 1.3500 s, value at t (y) = 13.4600$
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$y_{0} = 20.1806$
Step 6 — Check: returning y0 = 20.1806 to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y_{0} = 20.1806

Why the other options are there

40.3612 — kept a factor of two that cancels in the correct rearrangement.
10.0903 — dropped that same factor in the other direction.
22.1987 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Differential Equations

Example 5

First-order linear ODE solution — solve for initial value — Differential Equations (5)

temperature decay of a mass concrete pour Given value at time t (y) = 7.7000 mg/L; rate constant (k) = 0.7000 1/h; elapsed time (t) = 5.6000 h, determine the initial value (y_0) in mg/L.

Given

$value at time t (y) = 7.7000 mg/L$
$rate constant (k) = 0.7000 1/h$
$elapsed time (t) = 5.6000 h$

Find

initial value (y_0), in mg/L

Start with the thinking

The governing relation printed in this handbook section is First-order linear ODE solution.
Everything except y_0 is given, so isolate y_0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The ODE y' + k y = 0 has the exponential solution shown.

Step-by-step solution

Step 1 — State the governing relation:
$y(t) = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for y_0:
$y_{0} = y_0 = y e^{k t}$
Step 3 — List the givens: value at time t (y) = 7.7000 mg/L, rate constant (k) = 0.7000 1/h, elapsed time (t) = 5.6000 h.
Step 4 — Substitute the given values:
$y_{0} = y_0 = 7.7000 e^{0.7000 5.6000}$
Step 5 — Evaluate:
$y_{0} = 388.1\ \text{mg/L}$
Step 6 — Check: returning y_0 = 388.1 mg/L to
$y(t) = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y_{0} = 388.1\ \text{mg/L}

Why the other options are there

776.2 — kept a factor of two that cancels in the correct rearrangement.
194.0 — dropped that same factor in the other direction.
426.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations (First Order)

Example 6

First-order linear differential equation solution — solve for initial condition — Differential Equations (6)

An engineer solves a first-order differential equation for a cooling process. Given rate constant (k) = 1.1500 1/s; time (t) = 2.8000 s; solution value (y) = 15.6800, determine the initial condition (y0).

Given

$rate constant (k) = 1.1500 1/s$
$time (t) = 2.8000 s$
$solution value (y) = 15.6800$

Find

initial condition (y0)

Start with the thinking

The governing relation printed in this handbook section is First-order linear differential equation solution.
Everything except y0 is given, so isolate y0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
This first-order differential equation dy/dt = -k y has the exponential decay solution shown.

Figure 6 — schematic for First-order linear differential equation solution — solve for initial condition — Differential Equations (6)

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for y0:
$y_{0} = \dfrac{y}{e^{-kt}}$
Step 3 — List the givens: rate constant (k) = 1.1500 1/s, time (t) = 2.8000 s, solution value (y) = 15.6800.
Step 4 — Substitute the given values:
$y_{0} = \dfrac{15.6800}{e^{-k2.8000}}$
Step 5 — Evaluate:
$y_{0} = 392.4$
Step 6 — Check: returning y0 = 392.4 to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

y_{0} = 392.4

Why the other options are there

784.9 — kept a factor of two that cancels in the correct rearrangement.
196.2 — dropped that same factor in the other direction.
431.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations

Example 7

Exponential decay — solve for decay constant — Differential Equations (7)

A mathematics problem uses Exponential decay. Given initial value (y0) = 37.5000; time (t) = 0.2000 s; value at t (y) = 46.3900, determine the decay constant (k) in 1/s.

Given

$initial value (y_{0}) = 37.5000$
$time (t) = 0.2000 s$
$value at t (y) = 46.3900$

Find

decay constant (k), in 1/s

Start with the thinking

The governing relation printed in this handbook section is Exponential decay.
Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange the relation so that k stands alone on the left-hand side.
Step 3
$List the givens: initial value (y_{0}) = 37.5000, time (t) = 0.2000 s, value at t (y) = 46.3900$
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$k = -1.0637\ \text{1/s}$
Step 6 — Check: returning k = -1.0637 1/s to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

k = -1.0637\ \text{1/s}

Why the other options are there

-2.1274 — kept a factor of two that cancels in the correct rearrangement.
-0.5319 — dropped that same factor in the other direction.
-1.1701 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Differential Equations

Example 8

First-order linear ODE solution — solve for rate constant — Differential Equations (8)

chlorine residual decay in a storage tank Given value at time t (y) = 5.1000 mg/L; initial value (y_0) = 11.7000 mg/L; elapsed time (t) = 4.8000 h, determine the rate constant (k) in 1/h.

Given

$value at time t (y) = 5.1000 mg/L$
$initial value (y_0) = 11.7000 mg/L$
$elapsed time (t) = 4.8000 h$

Find

rate constant (k), in 1/h

Start with the thinking

The governing relation printed in this handbook section is First-order linear ODE solution.
Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
The ODE y' + k y = 0 has the exponential solution shown.

Step-by-step solution

Step 1 — State the governing relation:
$y(t) = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for k:
$k = k = \dfrac{\ln(y_0 / y)}{t}$
Step 3 — List the givens: value at time t (y) = 5.1000 mg/L, initial value (y_0) = 11.7000 mg/L, elapsed time (t) = 4.8000 h.
Step 4 — Substitute the given values:
$k = k = \dfrac{\ln(y_0 / 5.1000)}{4.8000}$
Step 5 — Evaluate:
$k = 0.1730\ \text{1/h}$
Step 6 — Check: returning k = 0.1730 1/h to
$y(t) = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

k = 0.1730\ \text{1/h}

Why the other options are there

0.3460 — kept a factor of two that cancels in the correct rearrangement.
0.0865 — dropped that same factor in the other direction.
0.1903 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations (First Order)

Example 9

First-order linear differential equation solution — solve for time — Differential Equations (9)

A student verifies the solution of a linear differential equation at a given time. Given initial condition (y0) = 6.0000; rate constant (k) = 1.1500 1/s; solution value (y) = 27.5800, determine the time (t) in s.

Given

$initial condition (y_{0}) = 6.0000$
$rate constant (k) = 1.1500 1/s$
$solution value (y) = 27.5800$

Find

time (t), in s

Start with the thinking

The governing relation printed in this handbook section is First-order linear differential equation solution.
Everything except t is given, so isolate t symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
This first-order differential equation dy/dt = -k y has the exponential decay solution shown.

Figure 9 — schematic for First-order linear differential equation solution — solve for time — Differential Equations (9)

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange symbolically for t:
$t = -\dfrac{\ln(y/y_0)}{k}$
Step 3 — List the givens: initial condition (y0) = 6.0000, rate constant (k) = 1.1500 1/s, solution value (y) = 27.5800.
Step 4 — Substitute the given values:
$t = -\dfrac{\ln(27.5800/y_0)}{1.1500}$
Step 5 — Evaluate:
$t = -1.3264\ \text{s}$
Step 6 — Check: returning t = -1.3264 s to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

t = -1.3264\ \text{s}

Why the other options are there

-2.6528 — kept a factor of two that cancels in the correct rearrangement.
-0.6632 — dropped that same factor in the other direction.
-1.4590 — rounded an intermediate value before the final step.

Reference: FE Handbook — Differential Equations

Example 10

Exponential decay — solve for time — Differential Equations (10)

A mathematics problem uses Exponential decay. Given initial value (y0) = 39.5000; decay constant (k) = 0.8000 1/s; value at t (y) = 20.0500, determine the time (t) in s.

Given

$initial value (y_{0}) = 39.5000$
$decay constant (k) = 0.8000 1/s$
$value at t (y) = 20.0500$

Find

time (t), in s

Start with the thinking

The governing relation printed in this handbook section is Exponential decay.
Everything except t is given, so isolate t symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mathematics items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$y = y_0 e^{-k t}$
Step 2 — Rearrange the relation so that t stands alone on the left-hand side.
Step 3 — List the givens: initial value (y0) = 39.5000, decay constant (k) = 0.8000 1/s, value at t (y) = 20.0500.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$t = 0.8476\ \text{s}$
Step 6 — Check: returning t = 0.8476 s to
$y = y_0 e^{-k t}$
reproduces the given quantities, and both sides carry the same units.

Answer:

t = 0.8476\ \text{s}

Why the other options are there

1.6952 — kept a factor of two that cancels in the correct rearrangement.
0.4238 — dropped that same factor in the other direction.
0.9323 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mathematics → Differential Equations