How many separable equations practice problems are on this page?

17, each with the full worked solution shown — no sign-up and nothing hidden behind a paywall.

Differential Equations · First-Order Equations

Separable Equations — Practice Problems and Worked Solutions

Separate variables and lock in the initial condition.

The formulas you need here

Separable equation

\frac{d y}{d x} = g (x) h (y) \Rightarrow \int \frac{d y}{h ( y )} = \int g (x) d x

What it says: Get all the y on one side and all the x on the other, then integrate both.
When to reach for it: The right-hand side factors cleanly into an x part times a y part.
Watch out: One constant of integration is enough for both sides, and dividing by h(y) can lose the constant solution h(y) = 0.

y^{'} = x y \Rightarrow ln ∣ y ∣ = \frac{x ^{2}}{2} + C

Exponentiate at the end to get y = Ae^{x²/2}.

17 worked problems

Every step is generated and checked by a computer algebra system.

1.
Solve the separable initial-value problem.
$y^{'} = 1 y, y (0) = 2$
Answer: $2 e^{x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 2 e^{x}$
2.
Solve the separable initial-value problem.
$y^{'} = 2 y, y (0) = 1$
Answer: $e^{2 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = e^{2 x}$
3.
Solve the separable initial-value problem.
$y^{'} = - 1 y, y (0) = 3$
Answer: $3 e^{- x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 3 e^{- x}$
4.
Solve the separable initial-value problem.
$y^{'} = 3 y, y (0) = - 2$
Answer: $- 2 e^{3 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = - 2 e^{3 x}$
5.
Solve the separable initial-value problem.
$y^{'} = - 2 y, y (0) = 1$
Answer: $e^{- 2 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = e^{- 2 x}$
6.
Solve the separable initial-value problem.
$y^{'} = \frac{1}{2} y, y (0) = 4$
Answer: $4 e^{\frac{x}{2}}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 4 e^{\frac{x}{2}}$
7.
Solve the separable initial-value problem.
$y^{'} = 1 y, y (0) = 5$
Answer: $5 e^{x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 5 e^{x}$
8.
Solve the separable initial-value problem.
$y^{'} = 2 y, y (0) = 3$
Answer: $3 e^{2 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 3 e^{2 x}$
9.
Solve the separable initial-value problem.
$y^{'} = - 1 y, y (0) = 4$
Answer: $4 e^{- x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 4 e^{- x}$
10.
Solve the separable initial-value problem.
$y^{'} = 4 y, y (0) = 1$
Answer: $e^{4 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = e^{4 x}$
11.
Solve the separable initial-value problem.
$y^{'} = - 3 y, y (0) = 2$
Answer: $2 e^{- 3 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 2 e^{- 3 x}$
12.
Solve the separable initial-value problem.
$y^{'} = \frac{1}{3} y, y (0) = 6$
Answer: $6 e^{\frac{x}{3}}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 6 e^{\frac{x}{3}}$
13.
Solve the separable initial-value problem.
$y^{'} = 3 y, y (0) = 1$
Answer: $e^{3 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = e^{3 x}$
14.
Solve the separable initial-value problem.
$y^{'} = - 2 y, y (0) = 5$
Answer: $5 e^{- 2 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 5 e^{- 2 x}$
15.
Solve the separable initial-value problem.
$y^{'} = 5 y, y (0) = 2$
Answer: $2 e^{5 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = 2 e^{5 x}$
16.
Solve the separable initial-value problem.
$y^{'} = 1 y, y (0) = - 3$
Answer: $- 3 e^{x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = - 3 e^{x}$
17.
Solve the separable initial-value problem.
$y^{'} = 2 y, y (0) = - 1$
Answer: $- e^{2 x}$
Step-by-step solution
1. $\frac{d y}{y} = k d x$
2. $ln ∣ y ∣ = k x + C$
3. $y = - e^{2 x}$

The formulas you need here

Separable equation

17 worked problems

More from Differential Equations