Differential-Equations · Unit 12 · Video 2 · Interactive Practice

Particular Plus Complementary: The Form of Every Solution

IKey Formulas

Formula	Name	What it says
$y = y_p + y_c = y_p + c_1y_1 + c_2y_2$	General solution of $L\,y = f$	$y_p$ is any one solution of the full equation; $y_c$ solves $L\,y = 0$
$L(y_p + c_1y_1 + c_2y_2) = L(y_p) + L(c_1y_1 + c_2y_2) = f + 0$	Claim 1 (linearity)	Every member of the family is a solution
$L(u - y_p) = L(u) - L(y_p) = f - f = 0$	Claim 2 (linearity)	Any solution $u$ differs from $y_p$ by a solution of $L\,y = 0$
$y = e^{-kt}\displaystyle\int_0^t f(\tilde t)\,e^{k\tilde t}\,d\tilde t + c\,e^{-kt}$	First-order case $y' + ky = f(t)$	$y_p$ plus $y_c = c\,e^{-kt}$ ; for $k > 0$ , steady state plus transient

Key Insight: "Particular" means any one. Two different particular solutions differ by a solution of the homogeneous equation, so either one, plus $y_c$ , gives the same family.

IIVisualization 1 — Every Member of the Family Solves the Equation

For $y'' + y = 3\sin 2x$ , each choice of particular solution and constants still has $y'' + y = 3\sin 2x$ .

Particular solution c₁ 1 c₂ 0

IIIVisualization 2 — Any Solution Minus $y_p$

Any solution $u$ of $y'' + y = 3\sin 2x$ minus $y_p = -\sin 2x$ is a solution of $y'' + y = 0$ .

u(0) 1 u′(0) 0

IVVisualization 3 — Steady State and Transient in First Order

For $y' + ky = 1$ , the starting value fades when $k > 0$ and takes over when $k < 0$ .

k 1

VQuiz Questions

Problem 1 · Assembling the General Solution

Given: $y_p = -2$ is a particular solution of $y'' - 4y = 8$ — find the general solution.

✅ Correct!

y_c

solves

y'' - 4y = 0

(roots

\pm 2

), and adding the one particular solution

-2

gives every solution.

❌ Not quite. That is only

y_c

; it solves

y'' - 4y = 0

, not

= 8

. Add the particular solution.

❌ Check the roots.

r^2 - 4 = 0

gives

r = \pm 2

(real), so

y_c

uses

e^{\pm 2x}

, not sines and cosines.

❌ Check by plugging in. For the constant

2

0 - 4 \cdot 2 = -8 \neq 8

. The given

y_p = -2

is the one that works.

Show solution

Step 1, $y_c$ : the reduced equation $y'' - 4y = 0$ has $r^2 - 4 = 0$ , $r = \pm 2$ , so $y_c = c_1e^{2x} + c_2e^{-2x}$ .

Step 2, $y_p$ : check $-2$ : $(-2)'' - 4(-2) = 8$ . ✓

y = -2 + c_1e^{2x} + c_2e^{-2x}.

Problem 2 · Two Solutions of the Same Equation

Given: $u$ and $v$ both solve $L\,y = f(x)$ , where $L$ is linear and $f \neq 0$ — which function must solve the homogeneous equation $L\,y = 0$ ?

✅ Correct!

L(u - v) = L(u) - L(v) = f - f = 0

— the key step of Claim 2.

❌ Not quite.

L(u + v) = f + f = 2f

, which is not

0

when

f \neq 0

❌ Not quite. Linearity says nothing about products:

L(uv)

is not

L(u)\,L(v)

in general.

❌ Not quite.

L(2u - v) = 2f - f = f

: this is another solution of the full equation, not of the homogeneous one.

Show solution

By linearity, $L(au + bv) = a\,L(u) + b\,L(v) = (a + b)\,f$ . This is $0$ exactly when $a + b = 0$ , i.e. for multiples of $u - v$ :

L(u - v) = f - f = 0.

So any two particular solutions differ by a member of $y_c$ .

Problem 3 · Fitting the Initial Values

Given: $y'' + y = x$ has the particular solution $y_p = x$ — find the solution with $y(0) = 1$ , $y'(0) = 0$ .

✅ Correct!

y(0) = 1

and

y'(0) = 1 - 0 - 1 = 0

, and it solves the full equation.

❌ Check the slope.

y_p = x

contributes slope

1

0

, so

y'(0) = 1

here. The constants must be fitted to the whole

y_p + y_c

, not to

y_c

alone.

❌ Not quite.

\cos x

solves

y'' + y = 0

, not

= x

. The particular solution has to be included before fitting.

❌ Check the slope. Its derivative at

0

1 + 0 + 1 = 2

, not

0

Show solution

General solution: $y = x + c_1\cos x + c_2\sin x$ , so $y' = 1 - c_1\sin x + c_2\cos x$ .

y(0) = c_1 = 1,\qquad y'(0) = 1 + c_2 = 0 \ \Rightarrow\ c_2 = -1.

Hence $y = x + \cos x - \sin x$ .

Problem 4 · When the Split Means Nothing

Given: $y' - 3y = 6$ — which statement describes its solutions as $t \to \infty$ ?

✅ Correct! Here

k = -3 < 0

y = -2 + c\,e^{3t}

, and

c\,e^{3t}

grows whenever

c \neq 0

❌ That is the $k > 0$ picture. In

y' + ky = 6

form,

k = -3

, so

y_c = c\,e^{3t}

grows instead of dying away.

❌ Not quite. The constant solution satisfies

0 - 3y = 6

, so it is

y = -2

; and the other solutions do not approach it.

❌ Not quite.

y = 0

is not even a solution (

0 - 0 \neq 6

), and

c\,e^{3t}

grows for

c \neq 0

Show solution

Write it as $y' + ky = 6$ with $k = -3$ . A constant $y_p$ satisfies $-3y_p = 6$ , so $y_p = -2$ ; and $y_c = c\,e^{-kt} = c\,e^{3t}$ .

y = -2 + c\,e^{3t}.

For $c \neq 0$ the exponential runs off to $\pm\infty$ and dominates; the initial condition never stops mattering, so there is no steady state.

Solved: 0 / 4

Particular Plus Complementary: The Form of Every Solution

IKey Formulas

IIVisualization 1 — Every Member of the Family Solves the Equation

IIIVisualization 2 — Any Solution Minus ypy_pyp​

IVVisualization 3 — Steady State and Transient in First Order

VQuiz Questions

IIIVisualization 2 — Any Solution Minus $y_p$