Differential-Equations · Unit 4 · Video 3 · Interactive Practice

The Reciprocal Substitution: $y' = y/x - y^2$ Becomes Linear in $v = 1/y$

IKey Formulas

Formula	Name	What it is for
$y' = \dfrac{y}{x} - y^2 \iff x\,y' - y = -x\,y^2$	Bernoulli, $n = 2$	$p(x) = \tfrac{1}{x}$ , $q(x) = -1$ ; the second form hides the pattern
$v = \dfrac{1}{y}, \quad v' = -\dfrac{1}{y^2}\,y'$	Reciprocal substitution	$v = y^{1-n}$ ; the derivative is in $x$ , so the chain rule keeps $y'$
$v' + \dfrac{v}{x} = 1 \;\Longrightarrow\; (x\,v)' = x$	The linear equation	Integrating factor $e^{\int dx/x} = x$ for $x > 0$
$y = \dfrac{2x}{x^2 + C_1}$ and $y = 0$	All the solutions	$C_1 = 2c$ from $v = \tfrac{x}{2} + \tfrac{c}{x}$ ; $y = 0$ is lost by dividing by $y^2$

Key Insight: Learn the steps, not the formula: divide by $y^n$ , set $v = y^{1-n}$ with the chain rule, solve the linear equation, then invert the whole fraction. The constant that was added on in $x\,v = \tfrac{x^2}{2} + c$ ends up inside the denominator of $y$ .

IIThe Substitution, Checked Point by Point

Along each curve $y = 2x/(x^2 + C_1)$ , $v = 1/y$ satisfies $v' + v/x = 1$ , with $v' = -y'/y^2$ , not $-1/y^2$ .

Curve Point x 2 v′ taken as

━ $C_1 = \tfrac{1}{4}$ ━ $C_1 = 1$ ━ $C_1 = 4$

$-1/y^2$ is $\dfrac{dv}{dy}$ ; the equation is about $\dfrac{dv}{dx} = \dfrac{dv}{dy}\cdot\dfrac{dy}{dx} = -\dfrac{1}{y^2}\,y'$ .

IIIWhere the Constant Goes

The same three solutions written in $x\,v$ , then $v$ , then $y$ : the constant starts added on and ends inside a denominator.

Step 1

━ $c = \tfrac{1}{8}$ , $C_1 = \tfrac{1}{4}$ ━ $c = \tfrac{1}{2}$ , $C_1 = 1$ ━ $c = 2$ , $C_1 = 4$

Step 1 — Integrate

(x\,v)' = x \;\Longrightarrow\; x\,v = \frac{x^2}{2} + c

The constant is added on: every curve is

x\,v = \tfrac{x^2}{2}

raised by

c

Step 2 — Divide, combine

v = \frac{x}{2} + \frac{c}{x} = \frac{x^2 + 2c}{2x}

Now

c

multiplies

1/x

, a solution of

v' + v/x = 0

Step 3 — Invert, rename

y = \frac{2x}{x^2 + 2c} = \frac{2x}{x^2 + C_1}, \qquad C_1 = 2c

Invert the whole fraction:

\dfrac{1}{x/2 + c/x} \ne \dfrac{2}{x} + \dfrac{x}{c}

IVAn Initial Condition Picks $C_1$

A starting point $(x_0, y_0)$ with $y_0 \ne 0$ fixes one $C_1$ in $y = 2x/(x^2 + C_1)$ ; on the $x$ -axis only $y = 0$ passes.

x₀ 1 y₀ 1

━ $C_1 = \tfrac{1}{4}$ ━ $C_1 = 1$ ━ $C_1 = 4$ ━ through $(x_0, y_0)$

Dividing by $y^2$ assumed $y \ne 0$ , so no value of $C_1$ produces $y = 0$ ; yet it solves $y' = y/x - y^2$ , and the complete answer is the family together with $y = 0$ .

VQuiz Questions

Problem 1 · Unmask the Bernoulli Form

Given: $x\,y' + 2y = x^3 y^2$ for $x > 0$ — find $p(x)$ , $q(x)$ and $n$ in the Bernoulli form $y' = p(x)\,y + q(x)\,y^n$ .

✅ Correct! Divide by

x

to get

y' + \tfrac{2}{x}\,y = x^2 y^2

, then move the

y

term right:

y' = -\tfrac{2}{x}\,y + x^2\,y^2

❌ Divide by $x$ first. The Bernoulli form has

y'

alone on the left. Here

y'

carries the coefficient

x

, so every term must be divided by

x

before

p

and

q

can be read off.

❌ Check the sign of $p$ . The term

\tfrac{2}{x}\,y

sits on the left beside

y'

; moving it to the right makes it

-\tfrac{2}{x}\,y

❌ Divide every term. Dividing by

x

turns

x^3 y^2

into

x^2 y^2

as well, so

q = x^2

Show solution

Divide through by $x$ (allowed, since $x > 0$ ):

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

Move the $y$ term to the right-hand side:

y' = -\frac{2}{x}\,y + x^2\,y^2

So $p(x) = -\dfrac{2}{x}$ , $q(x) = x^2$ and $n = 2$ , and the substitution is $v = y^{1-n} = 1/y$ , exactly as in the video.

Problem 2 · The Constant Lives in the Final Form

Given: $y' = \dfrac{y}{x} - y^2$ with $y(2) = \tfrac{1}{2}$ — find the solution.

✅ Correct!

\tfrac{1}{2} = \dfrac{4}{4 + C_1}

gives

4 + C_1 = 8

, so

C_1 = 4

: the curve whose peak

(\sqrt{C_1}, 1/\sqrt{C_1}) = (2, \tfrac{1}{2})

is the starting point itself.

❌ This meets $y(2) = \tfrac{1}{2}$ but does not solve the equation. It inverts

v = \tfrac{x}{2} + \tfrac{c}{x}

term by term, and the reciprocal of a sum is not the sum of the reciprocals. At

x = 2

it has

y' = -\tfrac{3}{4}

, while the equation demands

\tfrac{y}{x} - y^2 = \tfrac{1}{4} - \tfrac{1}{4} = 0

❌ Redo the last step. From

\tfrac{1}{2} = \dfrac{4}{4 + C_1}

, cross-multiplying gives

4 + C_1 = 8

, not

4 + C_1 = 2

; this curve has

y(2) = 2

❌ That is the video's curve for $y(1) = 1$ . At

x = 2

it gives

y = \tfrac{4}{5}

, not

\tfrac{1}{2}

Show solution

Substitute $x = 2$ , $y = \tfrac{1}{2}$ into the final form $y = \dfrac{2x}{x^2 + C_1}$ :

\frac{1}{2} = \frac{4}{4 + C_1} \;\Longrightarrow\; 4 + C_1 = 8 \;\Longrightarrow\; C_1 = 4

y = \frac{2x}{x^2 + 4}

Check. By the quotient rule $y' = \dfrac{2(x^2 + 4) - 2x\cdot 2x}{(x^2 + 4)^2} = \dfrac{8 - 2x^2}{(x^2 + 4)^2}$ , and

\frac{y}{x} - y^2 = \frac{2}{x^2 + 4} - \frac{4x^2}{(x^2 + 4)^2} = \frac{2(x^2 + 4) - 4x^2}{(x^2 + 4)^2} = \frac{8 - 2x^2}{(x^2 + 4)^2}\ \checkmark

Inverting $v = \tfrac{x}{2} + \tfrac{c}{x}$ term by term would give $\tfrac{2}{x} + \tfrac{x}{c}$ ; fitting it to $y(2) = \tfrac{1}{2}$ forces $c = -4$ and produces $y = \tfrac{2}{x} - \tfrac{x}{4}$ , a curve through the right point that is not a solution.

Problem 3 · The Whole Method, Once More

Given: $x\,y' + y = x^2 y^2$ for $x > 0$ with $y(1) = \tfrac{1}{2}$ — find the linear equation for $v = 1/y$ and the solution.

The linear equation for $v = 1/y$

The solution

✅ Correct! For

v' - \tfrac{v}{x} = -x

the integrating factor is

\tfrac{1}{x}

, giving

\left(\tfrac{v}{x}\right)' = -1

and

v = Cx - x^2

; then

v(1) = 2

fixes

C = 3

❌ The chain rule brings a minus sign.

v' = -\dfrac{y'}{y^2}

, so

\dfrac{y'}{y^2} = -v'

, not

+v'

❌ Multiply every term by $-1$ . From

-v' + \tfrac{v}{x} = x

you get

v' - \tfrac{v}{x} = -x

; the right-hand side changes sign too.

❌ Not quite. Divide by

x

, then by

y^2

\dfrac{y'}{y^2} + \dfrac{1}{x}\cdot\dfrac{1}{y} = x

. Now replace

\dfrac{y'}{y^2}

-v'

and

\dfrac{1}{y}

v

❌ The integrating factor is not always $x$ . Here

p = -\tfrac{1}{x}

, so

e^{\int p\,dx} = e^{-\ln x} = \tfrac{1}{x}

. Multiplying by

x

gives

x\,v' - v

, which is not

(x\,v)' = x\,v' + v

❌ Invert the whole expression. Inverting

v = Cx - x^2

term by term gives

\tfrac{1}{Cx} - \tfrac{1}{x^2}

, which meets

y(1) = \tfrac{1}{2}

C = \tfrac{2}{3}

but does not solve the equation: at

x = 1

it gives

xy' + y = 1

while

x^2 y^2 = \tfrac{1}{4}

. The reciprocal of a difference is not the difference of the reciprocals.

❌ Convert the initial condition too. The condition for

v

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

, not

\tfrac{1}{2}

; this curve has

y(1) = 2

Show solution

Bernoulli form. Divide by $x$ : $y' + \dfrac{y}{x} = x\,y^2$ , so $n = 2$ and $v = 1/y$ .

Substitute. Divide by $y^2$ and use $v' = -y'/y^2$ :

\frac{y'}{y^2} + \frac{1}{x}\cdot\frac{1}{y} = x \;\Longrightarrow\; -v' + \frac{v}{x} = x \;\Longrightarrow\; v' - \frac{v}{x} = -x

Integrating factor. $p = -\tfrac{1}{x}$ , so $e^{\int -dx/x} = e^{-\ln x} = \dfrac{1}{x}$ :

\frac{v'}{x} - \frac{v}{x^2} = -1 \;\Longrightarrow\; \left(\frac{v}{x}\right)' = -1 \;\Longrightarrow\; \frac{v}{x} = -x + C \;\Longrightarrow\; v = Cx - x^2

Initial condition. $v(1) = 1/y(1) = 2$ , so $C - 1 = 2$ and $C = 3$ .

Invert. $v = 3x - x^2$ is already a single expression, so

y = \frac{1}{3x - x^2}, \qquad 0 < x < 3

Check. $x\,y' + y = \dfrac{-x(3 - 2x) + (3x - x^2)}{(3x - x^2)^2} = \dfrac{x^2}{(3x - x^2)^2} = x^2 y^2\ \checkmark$

Problem 4 · Transfer to a Cubic Term

Given: $y' = y - y^3$ , a Bernoulli equation with $n = 3$ — find the linear equation for $v = y^{1-n} = y^{-2}$ , and the constant solution the method loses.

The linear equation for $v = y^{-2}$

The constant solution the method loses

✅ Correct!

v' = -2y^{-3}y'

turns

\dfrac{y'}{y^3} = \dfrac{1}{y^2} - 1

into

-\tfrac{1}{2}v' = v - 1

; then

v = 1 + Ce^{-2x}

gives

y = \pm\dfrac{1}{\sqrt{1 + Ce^{-2x}}}

, and dividing by

y^3

dropped only

y = 0

❌ Keep the factor from the chain rule.

\dfrac{d}{dx}\,y^{-2} = -2\,y^{-3}\,y'

, so

\dfrac{y'}{y^3} = -\tfrac{1}{2}v'

; the

2

n - 1

❌ Check the sign. The exponent

-2

brings down a negative factor:

v' = -2\,y^{-3}\,y'

, so

\dfrac{y'}{y^3} = -\tfrac{1}{2}v'

, with a minus.

❌ That is the $n = 2$ substitution. With

v = 1/y

the equation becomes

v' + v = 1/v

, still nonlinear; for

n = 3

the exponent is

1 - n = -2

❌ That one is in the family.

C = 0

gives

v = 1

, that is

y^2 = 1

, which covers both

y = 1

and

y = -1

❌ One constant solution is missing.

y = 0

solves

y' = y - y^3

, but

v = y^{-2}

is undefined there; the division by

y^3

assumed

y \ne 0

Show solution

Divide by $y^3$ (assuming $y \ne 0$ ):

\frac{y'}{y^3} = \frac{1}{y^2} - 1

With $v = y^{-2}$ the chain rule gives $v' = -2\,y^{-3}\,y'$ , so $\dfrac{y'}{y^3} = -\tfrac{1}{2}v'$ :

-\tfrac{1}{2}v' = v - 1 \;\Longrightarrow\; v' + 2v = 2

With integrating factor $e^{2x}$ : $\left(e^{2x}v\right)' = 2e^{2x}$ , so $e^{2x}v = e^{2x} + C$ and $v = 1 + Ce^{-2x}$ .

Back to $y$ : $y^2 = \dfrac{1}{1 + Ce^{-2x}}$ , so $y = \pm\dfrac{1}{\sqrt{1 + Ce^{-2x}}}$ . Taking $C = 0$ gives $y = \pm 1$ . No value of $C$ gives $y = 0$ , which solves the equation ( $0 = 0 - 0$ ) but was removed by the division.

Solved: 0 / 4

The Reciprocal Substitution: y′=y/x−y2y' = y/x - y^2y′=y/x−y2 Becomes Linear in v=1/yv = 1/yv=1/y

IKey Formulas

IIThe Substitution, Checked Point by Point

IIIWhere the Constant Goes

IVAn Initial Condition Picks C1C_1C1​

VQuiz Questions

The Reciprocal Substitution: $y' = y/x - y^2$ Becomes Linear in $v = 1/y$

IVAn Initial Condition Picks $C_1$