DE-BOOK Summary & Exercises

Section 3.3 Summary & Exercises

Summary of the Key Ideas.

Antiderivatives as Solutions to Differential Equations
- Finding the general antiderivative of a function, $f (x),$ is equivalent to finding the general solution to the differential equation
  $\frac{d y}{d x} = f (x) .$
Direct Integration
- Process of integrating both sides of a differential equation to directly solve for the dependent variable, $y .$

Exercises Exercises

General Solution.

Find the general solution for each of the following differential equations. Combine constants where appropriate.

1.

\frac{d y}{d x} = 2 x - 5

Answer. Answer

y = x^{2} - 5 x + C

2.

\frac{d}{d x} [y] = e^{2 x}

Answer. Answer

y = \frac{1}{2} e^{2 x} + C

3.

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

Answer. Answer

y = \frac{\sin (x)}{x} + \frac{C}{x}

4.

\frac{d y}{d t} = \frac{1}{t^{2}} + t

Answer. Answer

y = - \frac{1}{t} + \frac{1}{2} t^{2} + C

5.

\frac{d y}{d x} = x e^{x}

Answer. Answer

y = (x - 1) e^{x} + C

6.

{[\sin (x) \cdot y]}^{'} = \cos^{2} (x)

Answer. Answer

y = \frac{\sin (x)}{2} + \frac{C}{\sin (x)}

7.

y^{'} = \ln (x) + x^{2}

Answer. Answer

y = x \ln (x) - x + \frac{1}{3} x^{3} + C

8.

\frac{d}{d P} [e^{P} \cdot Q] = P

Answer. Answer

Q = e^{- P} (\frac{P}{e^{P}} + C)

Particular Solution.

Find the particular solution for each of the following differential equations with the given initial condition.

9.

\frac{d y}{d x} = 3 x^{2} + 2, y (1) = 4

Answer. Answer

y = x^{3} + 2 x + 1

10.

\frac{d}{d t} [y] = \sin (t), y (0) = 2

Answer. Answer

y = - \cos (t) + 3

11.

\frac{d}{d x} [x \cdot y] = e^{x}, y (1) = 0

Answer. Answer

y = \frac{e^{x}}{x} - \frac{e}{x}

12.

\frac{d y}{d x} = x^{3} - 4, y (2) = 1

Answer. Answer

y = \frac{x^{4}}{4} - 4 x + 5

13.

\frac{d R}{d h} = \frac{1}{h} + h, R (1) = 3

Answer. Answer

R = \ln | h | + \frac{h^{2}}{2} + \frac{5}{2}

14.

\frac{d}{d x} [e^{x} \cdot y] = \sin (x), y (0) = 2

Answer. Answer

y = e^{- x} (- \cos (x) + 3)

15.

y^{'} = \cos (x) + x^{2}, y (0) = - 2

Answer. Answer

y = \sin (x) + \frac{x^{3}}{3} - 2

16.

\frac{d}{d x} [\tan (x) \cdot y] = \sec^{2} (x), y (\frac{π}{4}) = 1

Answer. Answer

y = \frac{x \sec^{2} (x)}{\tan (x)} + \frac{π / 4 - 1}{\tan (x)}

17.

Attempt to apply direct integration to the differential equation

\frac{d y}{d x} = x + y .

Get to the point where it becomes clear that you cannot solve for

y

directly. What is the obstacle?

Answer. Answer

Integrating both sides gives

\begin{aligned} \int \frac{d y}{d x} d x & = \int (x + y) d x \\ y + C_{1} & = \int x d x + \int y d x \\ y + C_{1} & = \frac{1}{2} x^{2} + C_{2} + \int y d x \\ y - \int y d x & = \frac{1}{2} x^{2} + C_{2} - C_{1} \end{aligned}

Without knowing

y,

we cannot simplify

\int y d x .

So the obstacle is that we are unable to combine the these

y

variables into a single

y

on the left side.

Exercise Group.

Convert this idea from the integrating factor into some exercises for the direct integration method.

18.

Note 5.

In this chapter, we will study a solution technique that will help us solve differential equations that are first order and linear. Consider this (carefully selected) example.

\begin{matrix} (14) & x^{7} \cdot \frac{d y}{d x} + 7 x^{6} \cdot y = e^{x} \end{matrix}

Notice that the dependent variable is

y

and the independent variable is

x,

so we seek to find a formula for

y

in terms of

x .

The DE is also first order and linear.

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