Differential-Equations · Unit 4 · Video 2 · Interactive Practice

The Bernoulli Equation Is Linear in $v = y^{1-n}$ : Direct and Inverse Substitution

IKey Formulas

Formula	Name	What to check
$u = 1 - x^2$ vs. $x = \sin u$	Direct vs. inverse substitution	Direct: the new variable stands alone on the left. Inverse: an old variable does.
$y' = p(x)\,y + q(x)\,y^n$	Bernoulli equation	$n \ne 0, 1$ , and every term on the right carries $y$ or a power of $y$ : no term in $x$ alone, no constant
$v = y^{1-n}, \quad v' = (1-n)\,y^{-n}\,y'$	The direct substitution	After dividing by $y^n$ , the left side is $\dfrac{y'}{y^n} = \dfrac{v'}{1-n}$
$v' + (n-1)\,p(x)\,v = (1-n)\,q(x)$	Standard linear form	Integrating factor from $(n-1)\,p$ , then back to $y = v^{1/(1-n)}$ for $y > 0$

Key Insight: After dividing by $y^n$ , the right side is linear in $y^{1-n}$ , and the left side $y^{-n}\,y'$ is the derivative of that same quantity divided by $1 - n$ . Naming it $v$ leaves nothing nonlinear. The one thing the division can cost is $y = 0$ , which is a solution whenever $n > 0$ .

IIWhy the Power Must Be $1 - n$

Substituting $v = y^k$ into $y' = p\,y + q\,y^n$ gives $v' = kp\,v + kq\,v^m$ with $m = \frac{k + n - 1}{k}$ ; only $k = 1 - n$ makes $m = 0$ .

Exponent n Trial power k 1

IIIThe Same Solutions, Seen Through $v = y^{-2}$

$v = y^{-2}$ sends each nonzero solution of $y' = y - y^3$ to a solution $v = 1 + Ce^{-2x}$ of a linear equation.

Initial value y(0) 0.5

IVQuiz Questions

Problem 1 · Direct or Inverse?

Given: four substitutions, each bringing a new variable into a problem stated in old ones — which one is an inverse substitution?

✅ Correct! The old variable

x

stands alone on the left, set equal to an expression in the new variable

\theta

. Then

dx = 2\sec^2\theta\,d\theta

and

4 + x^2 = 4\sec^2\theta

, so the integral becomes

\int \tfrac12\,d\theta

❌ That one is direct. The new variable

u

stands alone on the left, equal to an expression in the old

x

, and

du = \frac{dx}{x}

is already sitting in the integrand.

❌ That one is direct. It is the Bernoulli substitution

v = y^{1-n}

with

n = 5

, and the new variable

v

stands alone on the left.

❌ Written this way it is direct: the new

w

stands alone. The same relation turned around,

y = wx

, is the inverse form, just as

T_1 = T/M

and

T = MT_1

were the two forms of the cooling-law scaling.

❌ Not quite. Look only at the variable standing alone on the left-hand side: new means direct, old means inverse.

Show solution

Classify each substitution by the one variable standing alone on the left-hand side: a new variable means direct, an old variable means inverse.

$u = \ln x$ : the new $u$ stands alone, so it is direct.
$v = y^{-4}$ : the new $v$ stands alone, so it is direct. It is the Bernoulli substitution for $n = 5$ , since $1 - n = -4$ .
$x = 2\tan\theta$ : the old $x$ stands alone, so it is inverse.
$w = y/x$ : the new $w$ stands alone, so it is direct; its inverse form is $y = wx$ .

The inverse form is the one to substitute with when the old variable is what fills the problem: every $x$ in the integrand is replaced by $2\tan\theta$ , and

\int \frac{dx}{4 + x^2} = \int \frac{2\sec^2\theta\,d\theta}{4 + 4\tan^2\theta} = \int \frac{2\sec^2\theta}{4\sec^2\theta}\,d\theta = \int \tfrac12\,d\theta

Problem 2 · The Sign Flip

Given: $y' = 3y + e^{x}\,y^{4}$ and the substitution $v = y^{-3}$ — find the linear equation for $v$ in standard form.

✅ Correct!

v' = -3\,y^{-4}\,y' = -3\,y^{-4}\left(3y + e^{x}y^{4}\right) = -9v - 3e^{x}

, and moving

-9v

to the left gives

v' + 9v = -3e^{x}

❌ Check the sign of the $v$ term. Moving

-9v

from the right-hand side to the left makes it

+9v

; the coefficient in standard form is

(n-1)\,p = 3 \cdot 3 = 9

❌ The chain-rule factor is missing.

v' = (1-n)\,y^{-n}\,y' = -3\,y^{-4}\,y'

, so the divided left side

y'/y^4

-v'/3

, not

v'

❌ Only one term got the factor. Multiplying

-\tfrac13 v' = 3v + e^{x}

-3

multiplies both terms on the right:

v' = -9v - 3e^{x}

❌ Not quite. Divide by

y^4

, replace

y'/y^4

-v'/3

and

y^{-3}

v

, then clear the factor and move the

v

term left.

Show solution

Here $n = 4$ , $p = 3$ , $q = e^{x}$ , and $1 - n = -3$ . Divide through by $y^4$ :

\frac{y'}{y^4} = 3\,y^{-3} + e^{x}

The chain rule gives $v' = -3\,y^{-4}\,y'$ , so $\dfrac{y'}{y^4} = -\dfrac{v'}{3}$ ; and $y^{-3} = v$ :

-\frac{v'}{3} = 3v + e^{x}

Multiply every term by $-3$ , then move the $v$ term to the left, where its sign flips:

v' = -9v - 3e^{x} \quad\Longrightarrow\quad v' + 9v = -3e^{x}

Check against the general form $v' + (n-1)\,p\,v = (1-n)\,q$ : $(n-1)\,p = 3 \cdot 3 = 9$ and $(1-n)\,q = -3e^{x}$ . ✓

Problem 3 · All Five Steps

Given: $y' = y + x\,y^{2}$ with $y(0) = \tfrac12$ — find $y(1)$ .

✅ Correct!

v = 1/y

satisfies

v' + v = -x

, so

v = 1 - x + Ce^{-x}

;

v(0) = 2

gives

C = 1

, and

y(1) = 1/(1 - 1 + e^{-1}) = e

❌ A sign slipped in the standard form. Multiplying

-v' = v + x

-1

negates both terms on the right:

v' = -v - x

, so

v' + v = -x

, not

v' - v = -x

❌ That is $v(1)$ , not $y(1)$ . The linear solve gives

v = 1 - x + e^{-x}

, so

v(1) = e^{-1}

; the last step converts back:

y(1) = 1/v(1)

❌ The initial condition was left in $y$ . It has to go through the substitution too:

v(0) = 1/y(0) = 2

, not

\tfrac12

❌ Not quite. Divide by

y^2

, set

v = 1/y

, solve the linear equation for

v

with

v(0) = 2

, then invert.

Show solution

1. Check the form. $p = 1$ , $q = x$ , $n = 2$ ; both terms on the right carry a power of $y$ .

2. Divide by $y^2$ .

\frac{y'}{y^2} = \frac{1}{y} + x

3. Set $v = y^{1-n} = y^{-1}$ . Then $v' = -y^{-2}\,y'$ , so the left side is $-v'$ :

-v' = v + x

4. Standard form and integrating factor. $v' + v = -x$ , so the integrating factor is $e^{x}$ and

(e^{x}v)' = -x\,e^{x} \quad\Longrightarrow\quad e^{x}v = (1 - x)\,e^{x} + C \quad\Longrightarrow\quad v = 1 - x + Ce^{-x}

The initial condition is converted too: $v(0) = 1/y(0) = 2$ , so $1 + C = 2$ and $C = 1$ .

5. Back to $y$ .

y = \frac{1}{1 - x + e^{-x}}, \qquad y(1) = \frac{1}{1 - 1 + e^{-1}} = e

The denominator $1 - x + e^{-x}$ is decreasing and still equals $e^{-1} > 0$ at $x = 1$ , so the solution exists on all of $[0, 1]$ .

Problem 4 · A Growth Model

Given: in the Solow growth model, capital per worker $k(t) > 0$ obeys $k' = s\,k^{\alpha} - \delta\,k$ , with constants $s > 0$ , $\delta > 0$ and $0 < \alpha < 1$ — find the general solution.

✅ Correct! With

n = \alpha

p = -\delta

and

q = s

, the substitution

v = k^{1-\alpha}

gives the linear equation

v' + (1-\alpha)\,\delta\,v = (1-\alpha)\,s

, so

v

relaxes to

s/\delta

at rate

(1-\alpha)\delta

and

k

to the steady state

(s/\delta)^{1/(1-\alpha)}

❌ That is $v$ , not $k$ . With

v = k^{1-\alpha}

the linear solve gives

v = \frac{s}{\delta} + Ce^{-(1-\alpha)\delta t}

; the last step converts back with

k = v^{1/(1-\alpha)}

❌ The chain-rule factor went missing.

v' = (1-\alpha)\,k^{-\alpha}\,k'

, so dividing by

k^{\alpha}

gives

\frac{v'}{1-\alpha} = s - \delta v

, and

v

decays at rate

(1-\alpha)\delta

, not

\delta

❌ Check the sign when the $v$ term moves left. Here

p = -\delta

, so

v' = (1-\alpha)s - (1-\alpha)\delta\,v

becomes

v' + (1-\alpha)\delta\,v = (1-\alpha)s

: the coefficient

(n-1)\,p = (\alpha - 1)(-\delta)

is positive, and

v

decays toward

s/\delta

instead of growing.

❌ Not quite. Read off

n = \alpha

p = -\delta

q = s

, set

v = k^{1-\alpha}

, and solve the linear equation for

v

before converting back.

Show solution

1. Check the form. $k' = -\delta\,k + s\,k^{\alpha}$ is a Bernoulli equation with $p = -\delta$ , $q = s$ and $n = \alpha$ , which is neither $0$ nor $1$ .

2–3. Divide by $k^{\alpha}$ and set $v = k^{1-\alpha}$ . Since $v' = (1-\alpha)\,k^{-\alpha}\,k'$ ,

\frac{k'}{k^{\alpha}} = -\delta\,k^{1-\alpha} + s \quad\Longrightarrow\quad \frac{v'}{1-\alpha} = -\delta\,v + s

4. Standard form. Multiply by $1 - \alpha$ and move the $v$ term to the left:

v' + (1-\alpha)\,\delta\,v = (1-\alpha)\,s

The integrating factor is $e^{(1-\alpha)\delta t}$ , so $\big(e^{(1-\alpha)\delta t}\,v\big)' = (1-\alpha)\,s\,e^{(1-\alpha)\delta t}$ , which integrates to $e^{(1-\alpha)\delta t}\,v = \dfrac{s}{\delta}\,e^{(1-\alpha)\delta t} + C$ :

v = \frac{s}{\delta} + Ce^{-(1-\alpha)\delta t}

5. Back to $k$ , which is positive:

k = \left(\frac{s}{\delta} + Ce^{-(1-\alpha)\delta t}\right)^{1/(1-\alpha)}

As $t \to \infty$ the exponential dies out and $k \to (s/\delta)^{1/(1-\alpha)}$ , the steady state where $s\,k^{\alpha} = \delta\,k$ . Because $n = \alpha > 0$ , $k = 0$ also solves the equation: it is the solution the division by $k^{\alpha}$ drops.

Solved: 0 / 4

The Bernoulli Equation Is Linear in v=y1−nv = y^{1-n}v=y1−n: Direct and Inverse Substitution

IKey Formulas

IIWhy the Power Must Be 1−n1 - n1−n

IIIThe Same Solutions, Seen Through v=y−2v = y^{-2}v=y−2

IVQuiz Questions

The Bernoulli Equation Is Linear in $v = y^{1-n}$ : Direct and Inverse Substitution

IIWhy the Power Must Be $1 - n$

IIIThe Same Solutions, Seen Through $v = y^{-2}$