DE-BOOK Solutions by Direct Integration

Section 3.2 Solutions by Direct Integration

As seen in the previous example, if a differential equation can be written in the form:

\frac{d}{d x} [y] = f (x),

then the solution can be found by integrating both sides. This principle extends to more complex equations of the form:

\frac{d}{d x} [g (x, y)] = f (x) .

Here,

f (x)

is a placeholder that indicates that only

x

variables can appear on the right-hand side of this equation. Similarly,

g (x, y)

indicates that only

x

y

variables can appear inside the derivative.

This leads us to our first method for solving first-order differential equations—direct integration.

Method 1. Direct Integration.

The general solution of a differential equation in the form

\begin{matrix} (13) & \frac{d}{d x} [g (x, y)] = f (x), \end{matrix}

can be determined by integrating both sides with respect to

x .

Consider the following examples to see how the same approach applies to these more interesting problems.

Example 3.

\frac{d}{d x} [5 x \cdot y] = \frac{1}{x^{2}}, y (1) = - 4 .

Solution. Solution

To solve for

y,

we need to eliminate the derivative by integrating both sides as before.

\begin{aligned} \int \frac{d}{d x} [5 x \cdot y] d x & = \int x^{- 2} d x \\ 5 x \cdot y + c_{1} & = - x^{- 1} + c_{2} \\ y & = \frac{1}{5 x} (- \frac{1}{x} + \overset{= c}{\overset{⏞}{c_{2} - c_{1}}}) \\ y & = \frac{1}{5 x} (- \frac{1}{x} + c) \end{aligned}

Finally, we use the condition,

y (1) = - 4,

to find the particular solution.

\begin{aligned} - 4 & = \frac{1}{5} (- 1 + c) \\ - 20 & = - 1 + c \\ c & = - 19 \end{aligned}

Therefore, the particular solution is

y = \frac{1}{5 x} (- \frac{1}{x} - 19) .

Example 4.

\frac{d}{d x} [y \sin (2 x)] = \cos x

Solution. Solution

To solve the differential equation

\frac{d}{d x} [y \sin (2 x)] = \cos x,

we integrate both sides with respect to

x

to get

\begin{aligned} \int \frac{d}{d x} [y \sin (2 x)] d x & = \int \cos x d x \\ y \sin (2 x) + c_{1} & = \sin x + c_{2} \end{aligned}

Finally, we isolate

y

and combine constants.

\begin{aligned} y \sin (2 x) & = \sin x + \underset{= c}{\underset{⏟}{c_{2} - c_{1}}} \\ y & = \frac{\sin x + c}{\sin (2 x)} \end{aligned}

Reading Questions Check-Point Questions

1. Solving a differential equation by direct integration involves computing a derivative.

Solving a differential equation by direct integration involves computing a derivative.

True
Incorrect, direct integration involves integrating both sides of the equation.
False
Correct! Direct integration involves integrating both sides of the equation.

2. Direct integration could be used to solve the equation $\frac{d}{d x} [y^{2} + x^{3}] = \sqrt{x}$ .

Direct integration could be used to solve the equation
$\frac{d}{d x} [y^{2} + x^{3}] = \sqrt{x} .$

True
Correct!
False
Incorrect. This equation is in the form (13).

3. Why shouldn’t direct integration be applied to $\frac{d}{d x} [\frac{x}{y^{2}}] = \sin (x + y)$ .

Why shouldn’t direct integration be applied to the equation
$\frac{d}{d x} [\frac{x}{y^{2}}] = \sin (x + y) .$

The equation is not linear.
Incorrect, direct integration can handle this.
The $y$ term is squared.
Incorrect, direct integration can handle this.
The $y$ term cannot be in the denominator.
Incorrect, direct integration can handle this.
The right-hand side contains $y .$
Correct! Direct integration only works when the right-hand side contains only $x .$

4. In the equation $\frac{d}{d x} [5 x \cdot y] = \frac{1}{x^{2}},$ what is the first step in solving for $y ?$

In the differential equation
$\frac{d}{d x} [5 x \cdot y] = \frac{1}{x^{2}},$
what is the first step in solving for $y ?$

Integrate both sides with respect to $x .$
Correct! Integrating both sides is the first step in solving for $y .$
Differentiate both sides with respect to $x .$
Incorrect. Differentiating would not help solve the equation; integration is the correct approach.
Factor the expression $5 x \cdot y .$
Incorrect. Factoring is not necessary here; integration is the correct step.
Substitute a new variable for $5 x \cdot y .$
Incorrect. Substitution is not needed; direct integration is the correct step.

5. Solve $y^{'} = e^{2 x} - 5 x$ using direct integration.

Complete each step below to solve the differential equation

y^{'} = e^{2 x} - 4 x .

The dependent variable is ,

The independent variable is ,
Give the general solution. Don’t forget the constant of integration.

$y (x) =$

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