Worked examples · Sheet 1

Separable equations

Theory used: Separable equation Method of separation of variables General solution and particular solution

Exercise 1

Find the general solution of

y^{'} = 3 x^{2} y .

The constant function $y \equiv 0$ is a solution. For $y \neq = 0$ :

\int \frac{d y}{y} = \int 3 x^{2} d x

ln ∣ y ∣ = x^{3} + c

∣ y ∣ = e^{c} e^{x^{3}} ⟹ y = C e^{x^{3}}, C \neq = 0.

For $C = 0$ one recovers the zero solution, so the general solution is $y = C e^{x^{3}}$ with $C \in R$ .

y = e^{x^{3}}

y = \dfrac{1}{2} e^{x^{3}}

y = -e^{x^{3}}

Three solutions

y = C e^{x^{3}}

C = 1

in color,

C = \frac{1}{2}

dashed,

C = - 1

in red; as

x \to - \infty

they all tend to

0

The equation has the form $y^{'} = g (x) h (y)$ with $g (x) = 3 x^{2}$ and $h (y) = y$ , so it is separable. The factor $h (y) = y$ vanishes for $y = 0$ , therefore the zero function is a constant solution and must be recorded before dividing. In the regions where $y \neq = 0$ one divides both sides by $y$ and integrates: the left-hand side with respect to $y$ , the right-hand side with respect to $x$ . From the equality $ln ∣ y ∣ = x^{3} + c$ one obtains $∣ y ∣ = e^{c} e^{x^{3}}$ , and removing the absolute value turns the constant $\pm e^{c}$ into a nonzero constant $C$ . Since the choice $C = 0$ gives back the zero solution, a single formula collects all the solutions.

Exercise 2

Find the general solution of

y^{'} = e^{x - y} .

y^{'} = e^{x} e^{- y} ⟹ \int e^{y} d y = \int e^{x} d x

e^{y} = e^{x} + C

y = ln (e^{x} + C), C \in R,

on every interval where $e^{x} + C > 0$ .

By the properties of powers, $e^{x - y}$ is the product $e^{x} e^{- y}$ , so the equation is separable with $g (x) = e^{x}$ and $h (y) = e^{- y}$ . The factor $e^{- y}$ never vanishes, therefore there are no constant solutions. Dividing by $e^{- y}$ , that is, multiplying by $e^{y}$ , separates the variables, and integrating both sides gives $e^{y} = e^{x} + C$ . The left-hand side is always positive, so the equality defines a solution only where $e^{x} + C$ is positive: for $C \geq 0$ on all of $R$ , for $C < 0$ on the half-line $x > ln (- C)$ . Applying the logarithm gives the explicit form of the solution.

Exercise 3

Find the general solution of

y^{'} = x (1 + y^{2}) .

\int \frac{d y}{1 + y ^{2}} = \int x d x

arctan y = \frac{x ^{2}}{2} + C

y = tan (\frac{x ^{2}}{2} + C),

on every interval where $\frac{x ^{2}}{2} + C$ stays within an interval of the form $(- \frac{π}{2} + k π, \frac{π}{2} + k π)$ .

The equation is separable with $g (x) = x$ and $h (y) = 1 + y^{2}$ . The factor $1 + y^{2}$ is always greater than or equal to $1$ , so there are no constant solutions and the division is legitimate everywhere. An antiderivative of $\frac{1}{1 + y ^{2}}$ is $arctan y$ , therefore integrating both sides gives $arctan y = \frac{x ^{2}}{2} + C$ . Applying the tangent gives the explicit form $y = tan (\frac{x ^{2}}{2} + C)$ , which by direct verification satisfies the equation on every interval where the tangent is defined. Since the tangent has vertical asymptotes at the points $\frac{π}{2} + k π$ , each solution is defined only on the intervals where the argument avoids these values.