Differential Equations and Separable Variables — Worked Solutions

Ques­tion 1: Order and degree#

The prob­lem#

State the order and degree of:

(a) $y''' + 2y'' - y' = 0$

(b) $(y')^2 + 3y = \cos x$

(c) $\left(1 + (y')^2\right)^{3/2} = y''$ (square both sides first)

Under­stand­ing the prob­lem#

The order is the order of the high­est deriv­a­tive present. The degree is the power of that high­est deriv­a­tive, but only once the equa­tion is a poly­no­mial in the deriv­a­tives (no frac­tional pow­ers, no deriv­a­tive inside $\sin$, etc.).

The idea#

Find the high­est deriv­a­tive, make sure the equa­tion is poly­no­mial in the deriv­a­tives, then read its power.

Step-by-step solu­tion#

Part (a)

Step 1. The high­est deriv­a­tive is $y'''$, so the order is $3$.

Step 2. $y'''$ appears to the power $1$, and every­thing is already poly­no­mial. Degree $1$.

Part (b)

Step 1. The only deriv­a­tive is $y'$, so the order is $1$.

Step 2. It appears as $(y')^2$, so the degree is $2$. (The $\cos x$ involves only $x$, not a deriv­a­tive, so it does not mat­ter.)

Part (c)

Step 1. The power $\displaystyle \tfrac{3}{2}$ is not a whole num­ber, so the equa­tion is not yet a poly­no­mial in the deriv­a­tives. Square both sides.

$$\left(1 + (y')^2\right)^3 = (y'')^2$$

Step 2. The high­est deriv­a­tive is $y''$: order $2$.

Step 3. $y''$ appears squared: degree $2$.

Check­ing the answer#

In (c) the left side, when expanded, con­tains $(y')^6$, but $y'$ is not the high­est deriv­a­tive, so its power does not decide the degree.

Answer#

(a) order $3$, degree $1$ (b) order $1$, degree $2$ (c) order $2$, degree $2$

Com­mon mis­take to avoid#

In (c), read­ing the degree as $1$ from the orig­i­nal form. Always clear frac­tional pow­ers first.

Ques­tion 2: Ver­i­fy­ing a solu­tion#

The prob­lem#

Ver­ify that $y = e^{-x} + 2x - 2$ sat­is­fies $y' + y = 2x$.

Under­stand­ing the prob­lem#

"Ver­ify" means: dif­fer­en­ti­ate the given $y$, sub­sti­tute $y$ and $y'$ into the left side, and show you get the right side.

The idea#

Direct sub­sti­tu­tion.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$y' = -e^{-x} + 2$$

Step 2. Add $y$.

$$\begin{aligned} y' + y &= (-e^{-x} + 2) + (e^{-x} + 2x - 2)\\ &= 2x \end{aligned}$$

The $e^{-x}$ terms can­cel and $2 - 2 = 0$.

Check­ing the answer#

At $x = 0$: $y = 1 + 0 - 2 = -1$, $y' = -1 + 2 = 1$; $y' + y = 0 = 2 \times 0$.

Answer#

$y' + y = 2x$, so the func­tion is a solu­tion.

Ques­tion 3: Ver­i­fy­ing a solu­tion with a square root#

The prob­lem#

Ver­ify that $y = \sqrt{1 + x^2}$ sat­is­fies $yy' = x$.

Under­stand­ing the prob­lem#

Find $y'$ by the chain rule, then mul­ti­ply by $y$.

The idea#

$\displaystyle \frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate with $u = 1 + x^2$.

$$\displaystyle y' = \frac{1}{2\sqrt{1 + x^2}}\cdot 2x = \frac{x}{\sqrt{1 + x^2}}$$

Step 2. Mul­ti­ply by $y$.

$$\displaystyle yy' = \sqrt{1 + x^2}\cdot\frac{x}{\sqrt{1 + x^2}} = x$$

Check­ing the answer#

At $x = 0$: $y = 1$, $y' = 0$, and $yy' = 0 = x$.

Answer#

$yy' = x$, so the func­tion is a solu­tion.

Ques­tion 4: Form­ing an equa­tion with two con­stants#

The prob­lem#

Form the dif­fer­en­tial equa­tion of $y = ae^{2x} + be^{-x}$.

Under­stand­ing the prob­lem#

$a$ and $b$ are arbi­trary con­stants. We need an equa­tion in $x$, $y$ and deriv­a­tives only, with no $a$ or $b$.

The idea#

Two con­stants means dif­fer­en­ti­ate twice, giv­ing three equa­tions; then elim­i­nate $a$ and $b$.

Step-by-step solu­tion#

Step 1. Write $y$ and its first two deriv­a­tives.

$$\begin{aligned} y &= ae^{2x} + be^{-x} \quad &(1)\\ y' &= 2ae^{2x} - be^{-x} \quad &(2)\\ y'' &= 4ae^{2x} + be^{-x} \quad &(3) \end{aligned}$$

Step 2. Look for num­bers $p$, $q$ with $y'' + py' + qy = 0$. Col­lect­ing the $ae^{2x}$ terms: $4 + 2p + q = 0$. Col­lect­ing the $be^{-x}$ terms: $1 - p + q = 0$.

Step 3. Solve. Sub­tract the sec­ond from the first: $3 + 3p = 0$, so $p = -1$. Then $q = p - 1 = -2$.

Step 4. Write the equa­tion.

$$y'' - y' - 2y = 0$$

Check­ing the answer#

Sub­sti­tute (1), (2), (3): $(4a - 2a - 2a)e^{2x} + (b + b - 2b)e^{-x} = 0$. Cor­rect.

Answer#

$y'' - y' - 2y = 0$

Ques­tion 5: Cir­cles touch­ing the $y$-axis at the ori­gin#

The prob­lem#

Form the dif­fer­en­tial equa­tion of all cir­cles touch­ing the $y$-axis at the ori­gin, $x^2 + y^2 = 2ax$.

Under­stand­ing the prob­lem#

Each cir­cle has cen­tre $(a, 0)$ and radius $|a|$, so it touches the $y$-axis at $(0, 0)$. There is one con­stant, $a$.

The idea#

Dif­fer­en­ti­ate once, express $a$ in terms of $x$, $y$, $y'$, and sub­sti­tute back into the orig­i­nal equa­tion.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate $x^2 + y^2 = 2ax$ with respect to $x$ (remem­ber $y$ is a func­tion of $x$).

$$2x + 2yy' = 2a \;\Rightarrow\; a = x + yy'$$

Step 2. Sub­sti­tute $a$ into the orig­i­nal equa­tion.

$$x^2 + y^2 = 2x(x + yy') = 2x^2 + 2xyy'$$

Step 3. Rearrange.

$$2xyy' = y^2 - x^2$$

Check­ing the answer#

Take the cir­cle $a = 1$: $(x - 1)^2 + y^2 = 1$. At the point $(1, 1)$ the tan­gent is hor­i­zon­tal, so $y' = 0$; the equa­tion gives $2 \cdot 1 \cdot 1 \cdot 0 = 1 - 1 = 0$. Cor­rect.

Answer#

$2xyy' = y^2 - x^2$

Ques­tion 6: Sep­a­rat­ing the vari­ables#

The prob­lem#

Solve:

(a) $\displaystyle \frac{dy}{dx} = \frac{1 + y}{1 + x}$

(b) $e^x\tan y\,dx + (1 - e^x)\sec^2 y\,dy = 0$

Under­stand­ing the prob­lem#

In each equa­tion we can move all the $y$ terms to one side with $dy$ and all the $x$ terms to the other side with $dx$. That is what "vari­ables sep­a­ra­ble" means.

The idea#

Sep­a­rate, inte­grate both sides, and add one con­stant.

Step-by-step solu­tion#

Part (a)

Step 1. Sep­a­rate.

$$\displaystyle \frac{dy}{1 + y} = \frac{dx}{1 + x}$$

Step 2. Inte­grate both sides.

$$\log|1 + y| = \log|1 + x| + \log|C|$$

(Writ­ing the con­stant as $\log|C|$ makes the next step neat.)

Step 3. Com­bine the log­a­rithms and remove them.

$$\log|1 + y| = \log|C(1 + x)| \;\Rightarrow\; 1 + y = C(1 + x)$$

Part (b)

Step 1. Move the $dy$ term to the other side.

$$e^x\tan y\,dx = -(1 - e^x)\sec^2 y\,dy = (e^x - 1)\sec^2 y\,dy$$

Step 2. Divide by $\tan y\,(e^x - 1)$ to sep­a­rate.

$$\displaystyle \frac{\sec^2 y}{\tan y}\,dy = \frac{e^x}{e^x - 1}\,dx$$

Step 3. Inte­grate. On each side the numer­a­tor is the deriv­a­tive of the denom­i­na­tor, so each inte­gral is a log­a­rithm.

$$\log|\tan y| = \log|e^x - 1| + \log|C|$$

Step 4. Remove the log­a­rithms.

$$\tan y = C(e^x - 1)$$

Since $C$ is any con­stant, you can equally write this as $\tan y = C(1 - e^x)$ (replace $C$ by $-C$).

Check­ing the answer#

(a) Dif­fer­en­ti­ate $y = C(1 + x) - 1$: $\displaystyle y' = C = \frac{1 + y}{1 + x}$. Cor­rect.

Answer#

(a) $1 + y = C(1 + x)$ (b) $\tan y = C(1 - e^x)$

Ques­tion 7: A par­tic­u­lar solu­tion of $\displaystyle \frac{dy}{dx} = 2xy$#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} = 2xy$ with $y(0) = 3$.

Under­stand­ing the prob­lem#

We need the gen­eral solu­tion first, then the par­tic­u­lar one that passes through $x = 0$, $y = 3$.

The idea#

Sep­a­rate: $\displaystyle \frac{dy}{y} = 2x\,dx$, inte­grate, then use the con­di­tion to find the con­stant.

Step-by-step solu­tion#

Step 1. Sep­a­rate and inte­grate.

$$\displaystyle \frac{dy}{y} = 2x\,dx \;\Rightarrow\; \log|y| = x^2 + c$$

Step 2. Remove the log­a­rithm.

$$y = Ae^{x^2} \quad (A = \pm e^c)$$

Step 3. Use $y(0) = 3$.

$$3 = Ae^{0} = A$$

Step 4. Write the par­tic­u­lar solu­tion.

$$y = 3e^{x^2}$$

Check­ing the answer#

$y' = 3e^{x^2}\cdot 2x = 2x\,y$, and $y(0) = 3$.

Answer#

$y = 3e^{x^2}$

Ques­tion 8: A par­tic­u­lar solu­tion of $\displaystyle \frac{dy}{dx} = \frac{y}{x}$#

The prob­lem#

Solve $\displaystyle \frac{dy}{dx} = \frac{y}{x}$ with $y(1) = 4$.

Under­stand­ing the prob­lem#

Sep­a­rate, inte­grate, then use $x = 1$, $y = 4$.

The idea#

$\displaystyle \frac{dy}{y} = \frac{dx}{x}$, which inte­grates to log­a­rithms.

Step-by-step solu­tion#

Step 1. Sep­a­rate and inte­grate.

$$\displaystyle \frac{dy}{y} = \frac{dx}{x} \;\Rightarrow\; \log|y| = \log|x| + \log|C|$$

Step 2. Remove the log­a­rithms.

$$y = Cx$$

Step 3. Use $y(1) = 4$: $4 = C \times 1$, so $C = 4$.

$$y = 4x$$

Check­ing the answer#

$y' = 4$ and $\displaystyle \frac{y}{x} = \frac{4x}{x} = 4$. Also $y(1) = 4$.

Answer#

$y = 4x$

Ques­tion 9: Radioac­tive decay#

The prob­lem#

A radioac­tive sub­stance decays at a rate pro­por­tional to the amount present; half dis­ap­pears in $50$ years. How long until $\displaystyle \tfrac{1}{8}$ remains?

Under­stand­ing the prob­lem#

Let $N$ be the amount at time $t$ years, with $N_0$ at $t = 0$. "Decays at a rate pro­por­tional to the amount" means $\displaystyle \frac{dN}{dt} = -kN$ with $k > 0$ (neg­a­tive because it decreases). "Half dis­ap­pears in $50$ years" means $\displaystyle N = \tfrac{1}{2}N_0$ when $t = 50$.

The idea#

Solve the sep­a­ra­ble equa­tion to get $N = N_0e^{-kt}$, find $k$ from the half-life, then find $t$ when $\displaystyle N = \tfrac{1}{8}N_0$.

Step-by-step solu­tion#

Step 1. Sep­a­rate and inte­grate.

$$\displaystyle \frac{dN}{N} = -k\,dt \;\Rightarrow\; \log N = -kt + c \;\Rightarrow\; N = N_0e^{-kt}$$

Step 2. Use the half-life.

$$\displaystyle \frac{1}{2}N_0 = N_0e^{-50k} \;\Rightarrow\; e^{-50k} = \frac{1}{2} \;\Rightarrow\; k = \frac{\log 2}{50}$$

Step 3. Set $\displaystyle N = \tfrac{1}{8}N_0$.

$$\displaystyle e^{-kt} = \frac{1}{8} \;\Rightarrow\; kt = \log 8 = 3\log 2$$

Step 4. Sub­sti­tute $k$.

$$\displaystyle t = \frac{3\log 2}{\log 2 / 50} = 3 \times 50 = 150$$

Check­ing the answer#

Think in half-lives: after $50$ years $\displaystyle \tfrac{1}{2}$ is left, after $100$ years $\displaystyle \tfrac{1}{4}$, after $150$ years $\displaystyle \tfrac{1}{8}$.

Answer#

$150$ years.

Ques­tion 10: A curve from its slope#

The prob­lem#

The slope of a curve at $(x, y)$ is $\displaystyle \frac{2y}{x}$, and the curve passes through $(1, 3)$. Find it.

Under­stand­ing the prob­lem#

"Slope at $(x, y)$" is $\displaystyle \frac{dy}{dx}$. So we must solve $\displaystyle \frac{dy}{dx} = \frac{2y}{x}$ with $y = 3$ when $x = 1$.

The idea#

Sep­a­rate the vari­ables, inte­grate, then use the point to fix the con­stant.

Step-by-step solu­tion#

Step 1. Sep­a­rate.

$$\displaystyle \frac{dy}{y} = \frac{2\,dx}{x}$$

Step 2. Inte­grate.

$$\log|y| = 2\log|x| + \log|C| = \log|Cx^2|$$

Step 3. Remove the log­a­rithms.

$$y = Cx^2$$

Step 4. Use the point $(1, 3)$: $3 = C \times 1^2$, so $C = 3$.

$$y = 3x^2$$

Check­ing the answer#

$\displaystyle \frac{dy}{dx} = 6x$ and $\displaystyle \frac{2y}{x} = \frac{6x^2}{x} = 6x$. Equal. And at $x = 1$, $y = 3$.

Answer#

The curve is $y = 3x^2$.