CE Board Exam Randomizer

⬅ Back to Differential Equations
📘 Key Concepts: Separable Equations

A separable differential equation can be rewritten so that all $y$-terms are on one side and all $x$-terms are on the other:

$$\frac{dy}{dx} = g(x)\,h(y) \quad\Longrightarrow\quad \frac{dy}{h(y)} = g(x)\,dx$$

Steps:

  1. Separate variables: move all $y$-terms (with $dy$) to the left, all $x$-terms (with $dx$) to the right.
  2. Integrate both sides.
  3. Solve for $y$ (explicit or implicit), then apply initial conditions to find $C$.

Common models:

Problem: Basic Separable Equation

Solve the differential equation $\dfrac{dy}{dx} = 2xy$ given the initial condition $y(0) = 3$.

DE – Separable – Problem 1 – Setup DE – Separable – Problem 1 – Diagram DE – Separable – Problem 1 – Diagram
Solution Solution Solution Solution

Problem: Trigonometric Separable Equation

Solve $\dfrac{dy}{dx} = y\cos x$ given $y(0) = e$.

DE – Separable – Problem 2 – Setup DE – Separable – Problem 2 – Diagram DE – Separable – Problem 2 – Diagram
Solution Solution Solution Solution

Problem: Population Growth

A bacterial colony numbers 500 at $t = 0$ and grows to 800 at $t = 2$ hours. Assuming exponential growth governed by $\dfrac{dP}{dt} = kP$:
a. Find the growth constant $k$.
b. Predict the population at $t = 5$ hours.

DE – Separable – Problem 3 – Setup DE – Separable – Problem 3 – Diagram DE – Separable – Problem 3 – Diagram
Solution Solution Solution Solution

Problem: Implicit Solution

Solve $(1 + x^2)\,dy = xy\,dx$ given $y(0) = 2$. Express the solution explicitly for $y$.

DE – Separable – Problem 4 – Setup DE – Separable – Problem 4 – Diagram DE – Separable – Problem 4 – Diagram
Solution Solution Solution Solution

Problem: Simple Decay by Separation

Solve dy/dx = -3y with y(0) = 10.

$$\frac{dy}{y}=-3\,dx \Rightarrow \ln|y|=-3x+C$$
$$y=Ce^{-3x},\qquad y(0)=10 \Rightarrow C=10$$

Answer: $y=10e^{-3x}$.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t1791

MSTE - Differential Equations / Elimination of Arbitrary Constants / BEMz

Determine the differential equation of a family of lines passing thru (h, k).

  1. $(y-k) dx - (x-h) dy = 0$
  2. $(x-h) + (y-k) = dy/dx$
  3. $(x-h) dx - (y-k) dy = 0$
  4. $(x+h) dx - (y-k) dy = 0$
A family of lines passing through $(h,k)$ is
$$y-k=m(x-h)$$
Differentiate:
$$dy=m\,dx$$
Since
$$m=\frac{y-k}{x-h}$$
then
$$dy=\frac{y-k}{x-h}dx$$
Rearrange:
$$(y-k)dx-(x-h)dy=0$$
Therefore, the differential equation is $\boxed{(y-k)dx-(x-h)dy=0}$.

Question Bank: t1792

MSTE - Differential Equations / Elimination of Arbitrary Constants / BEMz

What is the differential equation of the family of parabolas having their vertices at the origin and their foci on the x-axis

  1. $2x dy - y dx = 0$
  2. $x dy + y dx = 0$
  3. $2y dx - x dy = 0$
  4. $dy/dx - x = 0$
A parabola with vertex at the origin and focus on the x-axis has form
$$y^2=4ax$$
Differentiate:
$$2y\,dy=4a\,dx$$
From the original equation,
$$4a=\frac{y^2}{x}$$
Substitute:
$$2y\,dy=\frac{y^2}{x}dx$$
Divide by $y$ and rearrange:
$$2x\,dy-y\,dx=0$$
Therefore, the differential equation is $\boxed{2xdy-ydx=0}$.

Question Bank: t1793

MSTE - Differential Equations / Elimination of Arbitrary Constants / BEMz

Find the differential equations of the family of lines passing through the origin.

  1. $ydx - xdy = 0$
  2. $xdy - ydx = 0$
  3. $xdx + ydy = 0$
  4. $ydx + xdy = 0$
A family of lines through the origin has equation
$$y=Cx$$
Differentiate:
$$dy=C\,dx$$
Since $C=y/x$,
$$dy=\frac{y}{x}dx$$
$$x\,dy-y\,dx=0$$
Therefore, the differential equation is $\boxed{xdy-ydx=0}$. The first listed choice is the same equation multiplied by $-1$.

Question Bank: t1799

MSTE - Differential Equations / Variable Separable / BEMz

From the given differential equation $xdx+6y^{5dy} = 0$ solve for the constant of integration when x = 0, y = 2.

  1. $27x dx + 4y^{2} dy = 0$
  2. 58
  3. 48
  4. 64
Interpreting the scanned equation as
$$x\,dx+6y^5\,dy=0$$
Integrate:
$$\int x\,dx+\int 6y^5\,dy=C$$
$$\frac{x^2}{2}+y^6=C$$
Using $x=0$ and $y=2$:
$$C=0+2^6=64$$
Therefore, the constant of integration is $\boxed{64}$.

Question Bank: t1800

MSTE - Differential Equations / Variable Separable / BEMz

Find the equation of the curve which passes through points (1, 4) and (0, 2) if $d^{2}$ y/ $dx^{2} = 1$

  1. $2y = x^{2} + 3x + 4$
  2. $4y = 2x^{2} + x + 4$
  3. $5y = x^{2} + 2x + 2$
  4. $3y = x^{2} + x + 4$
Given
$$\frac{d^2y}{dx^2}=1$$
Integrate twice:
$$\frac{dy}{dx}=x+C_1$$
$$y=\frac{x^2}{2}+C_1x+C_2$$
Using $(0,2)$ gives $C_2=2$. Using $(1,4)$:
$$4=\frac{1}{2}+C_1+2$$
$$C_1=\frac{3}{2}$$
Thus,
$$y=\frac{x^2}{2}+\frac{3x}{2}+2$$
or
$$\boxed{2y=x^2+3x+4}$$

Question Bank: t1805

MSTE - Differential Equations / Variable Separable / BEMz

Determine the general solution of xdy + $ydx=0$.

  1. $xy = c$
  2. $\ln xy = c$
  3. $\ln x + \ln y = c$
  4. $x + y = c$
Recognize the differential of a product:
$$d(xy)=x\,dy+y\,dx$$
Given
$$x\,dy+y\,dx=0$$
so
$$d(xy)=0$$
Integrating,
$$xy=C$$
Therefore, the general solution is $\boxed{xy=c}$.