Worked examples · Sheet 3

First-order linear equations

Theory used: First-order linear equation Solution formula with the integrating factor Structure of the general solution

Exercise 1

Find the general solution of

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

Here $a (x) = 1$ , so $A (x) = x$ and the integrating factor is $e^{x}$ :

(e^{x} y)^{'} = e^{x} (y^{'} + y) = e^{2 x}

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

y = \frac{e ^{x}}{2} + C e^{- x}, C \in R .

The equation is linear with $a (x) = 1$ and $b (x) = e^{x}$ , so one multiplies by the integrating factor $e^{A (x)}$ with $A$ an antiderivative of $a$ , that is, by $e^{x}$ . After the multiplication, the left-hand side is the derivative of the product $e^{x} y$ , therefore the equation can be integrated directly. An antiderivative of $e^{2 x}$ is $\frac{e ^{2 x}}{2}$ , and dividing by $e^{x}$ gives the general solution. The term $C e^{- x}$ is the general solution of the associated homogeneous equation $y^{'} + y = 0$ , while $\frac{e ^{x}}{2}$ is a particular solution of the complete equation, in accordance with the structure theorem.

Exercise 2

Find the general solution of

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

Here $a (x) = 2 x$ , so $A (x) = x^{2}$ and the integrating factor is $e^{x^{2}}$ :

(e^{x^{2}} y)^{'} = x e^{x^{2}}

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

y = \frac{1}{2} + C e^{- x^{2}}, C \in R .

y = \dfrac{1}{2}

y = \dfrac{1}{2} + e^{-x^{2}}

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

The constant solution

y = \frac{1}{2}

(in color) and the solutions with

C = 1

(dashed) and

C = - 1

(in red): far from the origin the term

C e^{- x^{2}}

becomes negligible

The coefficient $a (x) = 2 x$ is not constant, so the integrating factor contains its antiderivative $x^{2}$ . Multiplying by $e^{x^{2}}$ , the left-hand side becomes the derivative of $e^{x^{2}} y$ and the right-hand side becomes $x e^{x^{2}}$ . This function is integrated by recognizing that the derivative of $e^{x^{2}}$ is $2 x e^{x^{2}}$ , therefore an antiderivative of $x e^{x^{2}}$ is $\frac{e ^{x^{2}}}{2}$ . Dividing by $e^{x^{2}}$ gives the general solution, the sum of the constant particular solution $\frac{1}{2}$ and the general solution of the homogeneous equation $C e^{- x^{2}}$ . The constant $y \equiv \frac{1}{2}$ can also be verified directly, since it has zero derivative and $2 x \cdot \frac{1}{2} = x$ .

Exercise 3

Find the general solution, for $x > 0$ , of

y^{'} - \frac{y}{x} = x^{2} .

Here $a (x) = - \frac{1}{x}$ , so $A (x) = - ln x$ and the integrating factor is $e^{- l n x} = \frac{1}{x}$ :

(\frac{y}{x})^{'} = \frac{1}{x} (y^{'} - \frac{y}{x}) = x

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

y = \frac{x ^{3}}{2} + C x, C \in R .

For $x > 0$ the coefficients $a (x) = - \frac{1}{x}$ and $b (x) = x^{2}$ are continuous, so the solution formula applies on this half-line. An antiderivative of $- \frac{1}{x}$ is $- ln x$ , therefore the integrating factor is $e^{- l n x}$ , that is, $\frac{1}{x}$ . After the multiplication, the left-hand side is the derivative of the quotient $\frac{y}{x}$ , as one checks with the product rule applied to $y \cdot \frac{1}{x}$ . Integration gives $\frac{y}{x} = \frac{x ^{2}}{2} + C$ , and multiplying by $x$ gives the general solution: the term $C x$ solves the associated homogeneous equation, the term $\frac{x ^{3}}{2}$ is a particular solution of the complete equation.