Expert answers to "Solving simple differential equation (t+4)dx=4(1+x2)dt"

Name: Solving simple differential equation (t+4)dx=4(1+x2)dt
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solving simple differential equation (t+4)dx=4(1+x2)dt

killjoy1990xb9

Answered question

2022-01-20

Solving simple differential equation

(t + 4) d x = 4 (1 + x^{2}) d t

Answer & Explanation

Bob Huerta

Beginner2022-01-20Added 41 answers

As you rearranged it in your post, we have

\int \frac{d x}{1 + x^{2}} = 4 \int \frac{d t}{t + 4}

As you discovered, the LHS is arctan. To find the value of the RHS, we may say

t + 4 = u so d u = d t

. Thus the RHS can be rewritten as

4 \int \frac{d u}{u} = 4 \log | u | + C_{2}

Because we want this integral to be in respect to t, we have

u = t + 4

, so

\arctan x + C_{1} = 4 \log | t + 4 | + C_{2}

To confirm that the integral we solved is indeed correct, you can simply differentiate

4 \log | t + 4 | + C_{2}

Thus

x = \tan (4 \log | t + 4 | + C_{2} - C_{1})

and

t = \exp (\frac{1}{4} (\arctan x + C_{1} - C_{2})) - 4

Dawn Neal

Beginner2022-01-21Added 35 answers

Here are some proceeding steps:
- Seperating the variables you get

\frac{d x}{1 + x^{2}} = 4 \cdot \frac{d t}{t + 4}

- Integrating both sides you have:

\tan^{- 1} (x) + C_{1} = 4 \cdot \log | t + 4 | + C_{2}

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