Single-Variable-Calculus · Unit 13 · Video 2 · Interactive Practice

General Solutions: Signs, Constants and Initial Conditions

IKey Formulas

Formula	Name	Meaning
$\displaystyle\int \frac{dy}{y} = \ln\|y\| + c$	Log antiderivative	Valid for $y < 0$ as well as $y > 0$ , so for every $y \ne 0$
$\|y\| = e^{c}e^{-x^2/2} \;\Rightarrow\; y = \pm e^{c}e^{-x^2/2}$	Both signs	$\pm e^{c}$ takes every nonzero value
$y = a\,e^{-x^2/2}$ , $a$ any constant	General solution of $\dfrac{dy}{dx} = -xy$	$a = \pm e^{c}$ for $a \ne 0$ ; $a = 0$ is the lost solution $y = 0$
$y(0) = 3 \;\Rightarrow\; a\,e^{0} = 3 \;\Rightarrow\; a = 3$	Initial condition	One value of $y$ picks one curve from the family

Key Insight: Dividing by $y$ throws away the solutions where $y = 0$ . Check them directly and add them back. The one constant from antidifferentiating starts out added ( $+\,c$ ) and becomes a factor ( $e^{c}$ ) after exponentiating.

IIVisualization 1 — Where Every Solution Comes From

The assumption $y > 0$ yields only the curves above the axis; $\ln|y|$ and the zero check supply the rest.

Step 1

Step 1 — Assume $y > 0$

\ln y = -\tfrac{x^2}{2} + c

y = e^{c}e^{-x^2/2},\quad e^{c} > 0

IIIVisualization 2 — One Value Picks One Curve

A single point $y(x_0) = y_0$ fixes $a = y_0\,e^{x_0^2/2}$ , and the curve keeps the sign of $a$ everywhere.

x₀ 0 y₀ 3

IVVisualization 3 — Added Constant, Then a Factor

The same $c$ shifts $\ln|y|$ up by $c$ and stretches $|y|$ by the factor $e^{c}$ .

Constant c 0.70

💡 Two constants $\ln|y| + c_1 = -\tfrac{x^2}{2} + c_2$ give nothing new: $c_2 - c_1$ is one arbitrary constant $c$ .

VQuiz Questions

Problem 1 · Pick the Curve

Given: $\dfrac{dy}{dx} = -xy$ with $y(0) = -2$ — find the solution.

✅ Correct!

a\,e^{0} = -2

gives

a = -2

; since

e^{-x^2/2} > 0

, this solution stays negative for every

x

❌ Close, but check the sign.

2e^{-x^2/2}

is a solution, but its value at

0

2

, not

-2

. Negative

a

is allowed:

a = \pm e^{c}

❌ Not quite. It does pass through

(0, -2)

, but the constant is a factor, not an added term:

y' = -xe^{-x^2/2}

while

-xy = -xe^{-x^2/2} + 3x

❌ Not quite. The exponent sign is off:

\tfrac{d}{dx}e^{x^2/2} = xe^{x^2/2}

, which solves

y' = xy

, not

y' = -xy

❌ Not quite. Put

x = 0

into

y = a\,e^{-x^2/2}

and solve for

a

Show solution

The general solution is $y = a\,e^{-x^2/2}$ with $a$ any constant. Setting $x = 0$ :

a\,e^{-0^2/2} = a \cdot 1 = -2 \quad\Longrightarrow\quad a = -2.

So $y = -2e^{-x^2/2}$ , the mirror image of the $a = 2$ bell curve. It solves the equation because $y' = -2(-x)e^{-x^2/2} = 2xe^{-x^2/2} = -x\,y$ .

Problem 2 · The Lost Solution

Given: $\dfrac{dy}{dx} = x(y - 2)$ is separated as $\dfrac{dy}{y - 2} = x\,dx$ , giving $\ln|y - 2| = \tfrac{x^2}{2} + c$ — find the solution that this division loses.

✅ Correct! The division by

y - 2

is impossible where

y = 2

, and the constant

y = 2

satisfies

0 = x(2 - 2)

❌ Not quite. Here the divisor is

y - 2

, not

y

. And

y = 0

is not a solution:

0 \ne x(0 - 2) = -2x

❌ Close, but…

\ln|y - 2|

recovers both signs of

y - 2

, yet it is defined only for

y \ne 2

, so the case

y - 2 = 0

is still missing.

❌ Not quite. A division loses the solutions that make the divisor zero; check that candidate in the original equation.

Show solution

Dividing by $y - 2$ requires $y \ne 2$ . Test the constant function $y = 2$ :

\frac{d}{dx}(2) = 0, \qquad x(2 - 2) = 0 \quad\checkmark

So $y = 2$ is a solution the method missed. Exponentiating gives $y - 2 = \pm e^{c}e^{x^2/2}$ , and adding back $y = 2$ as $a = 0$ gives the general solution

y = 2 + a\,e^{x^2/2}, \quad a \text{ any constant}.

Problem 3 · Family, Then One Member

Given: $\dfrac{dy}{dx} = 3x^2y$ with $y(1) = 5$ — find the general solution and the constant fixed by the initial condition.

What is the general solution?

What is $a$ ?

✅ Correct!

y = \tfrac{5}{e}e^{x^3} = 5e^{x^3 - 1}

, which equals

5

x = 1

❌ Not quite. The constant is additive only right after antidifferentiating:

\ln|y| = x^3 + c

. Exponentiating turns it into a factor.

❌ Check the general solution. Separate:

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

, and an antiderivative of

3x^2

x^3

❌ Close, but…

a\,e^{1} = 5

means dividing by

e

, not multiplying.

❌ Check $a$ . The initial value is at

x = 1

, not

x = 0

: substitute

x = 1

into

y = a\,e^{x^3}

Show solution

Separate (for $y \ne 0$ ) and integrate:

\frac{dy}{y} = 3x^2\,dx \quad\Longrightarrow\quad \ln|y| = x^3 + c \quad\Longrightarrow\quad y = \pm e^{c}e^{x^3}.

The lost solution $y = 0$ satisfies $0 = 3x^2 \cdot 0$ , so with $a = \pm e^{c}$ or $a = 0$ the general solution is $y = a\,e^{x^3}$ , $a$ any constant.

Initial condition:

y(1) = a\,e^{1} = 5 \quad\Longrightarrow\quad a = \frac{5}{e}, \qquad y = 5e^{x^3 - 1}.

Problem 4 · An Implicit Answer and Its Sign

Given: $\dfrac{dy}{dx} = \dfrac{4x}{y}$ with $y(0) = -3$ . Separating gives $y\,dy = 4x\,dx$ , so $\tfrac{y^2}{2} = 2x^2 + c$ — find the solution $y(x)$ .

✅ Correct! The implicit answer

y^2 = 4x^2 + 9

allows both signs, and

y(0) = -3

selects the negative branch.

❌ Close, but check the sign.

y^2 = 4x^2 + 9

gives

y = \pm\sqrt{4x^2 + 9}

; the positive root has

y(0) = 3

, not

-3

❌ Not quite.

\sqrt{4x^2 + 9} \ne 2x + 3

: a square root does not split over a sum. Also

(2x - 3)' = 2

, while

\tfrac{4x}{y} = \tfrac{4x}{2x - 3}

❌ Not quite. That has the shape of the

y' = 4xy

solutions; here

4x

is divided by

y

, so the separated form is

y\,dy = 4x\,dx

, with no logarithm.

❌ Not quite. Fix

c

from

y(0) = -3

first, then solve

y^2 = 4x^2 + 2c

for

y

Show solution

Integrating both sides of $y\,dy = 4x\,dx$ :

\frac{y^2}{2} = 2x^2 + c \quad\Longrightarrow\quad y^2 = 4x^2 + 2c.

At $x = 0$ , $y = -3$ : $9 = 0 + 2c$ , so $2c = 9$ and $y^2 = 4x^2 + 9$ . This implicit formula describes two curves, $y = \pm\sqrt{4x^2 + 9}$ . The starting value is negative, so

y = -\sqrt{4x^2 + 9}.

Check: $y' = -\dfrac{4x}{\sqrt{4x^2 + 9}} = \dfrac{4x}{-\sqrt{4x^2 + 9}} = \dfrac{4x}{y}$ ✓, and $y(0) = -\sqrt{9} = -3$ ✓.

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