Solve dy/dx = 2e^(3x) + 1, y(0) = 2 Step by Step

Step by step

The question

Question: solve dy/dx = 2e^(3x) + 1, with y(0) = 2.

dydx = 2e^(3x) + 1

y(0) = 2 is the starting value: when x is 0, y is 2.

y(0) = 2

1 The idea

dy/dx is only about x.

So y is a function whose derivative is 2e^(3x) + 1: an antiderivative.

y′ = 2e^(3x) + 1: y is a function whose derivative is 2e^(3x) + 1

2 Integrate both sides

Integrate both sides.

The left side, the integral of dy/dx, is just y.

y = ∫ (2e^(3x) + 1) dx

The rules, one for each kind of term.

∫ xⁿ dx = x^{n + 1}n + 1

∫ e^kx dx = e^kxk

Each term by its rule, and one + C: the number we cannot see after a derivative.

y = 2 × e^(3x)3 + x1 + C

Tidy the numbers: y = 2/3 × e^(3x) + x + C.

y = 23e^(3x) + x + C

Careful!

Don't forget + C.

Every number C gives another solution, and we need C to use the starting value.

Careful: not y = 23e^(3x) + x ✗

3 Use the starting value

Put in x = 0 and y = 2.

2 = 23e^(0) + (0) + C

e^(0) = 1

Work out the numbers: 2/3.

2 = 23 + C

To get C alone, take away 2/3 on both sides.

2 − 23 = C

So C = 4/3.

C = 43

Put C back.

The answer: y = 2/3 × e^(3x) + x + 4/3.

y = 23e^(3x) + x + 43

4 Check

Let's check: the derivative of our answer is 2e^(3x) + 1.

The right side again.

Right!

y′ = 2e^(3x) + 1

And at x = 0, y is 2: the starting value.

Right!

y(0) = 2

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer