0
$\begingroup$

I am trying to solve the following equation:

$(t+4)dx=4(1+x^2)dt$

As far as I can remember I have to move x to left and t to right and form an integration. What should I do? Does deviding both side by $(t+4)$ helps? what will be next?!

update

ok as mrf suggested I came to this:

$\int \frac 1{1+x^2}\,dx = 4\int\frac 1{t+4}\,dt$

This will be:

$\tan^{-1}(x)+C == $

I am not sure about right side...any tips?!

  • 2
    Remember... $\frac{d}{dx} 4\log |t+4| = \frac{4}{t+4}$2012-05-02

2 Answers 2

1

As you rearranged it in your post, we have

$\int \frac{dx}{1+x^2} = 4\int \frac{dt}{t+4}$

As you discovered, the LHS is $\arctan$. To find the value of the RHS, we may say $t+4=u$ so $du=dt$. Thus the RHS can be rewritten as

$4\int \frac{du}{u} = 4\log |u|+C_2$

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

$t = \exp (\frac{1}{4}(\arctan x+C_1-C_2))-4$

  • 0
    @Sean87 I am simply isolating $x$ and $t$ from $\arctan x+C_1=4\log|t+4|+C_2 \implies \arctan x=4\log|t+4|+C_2-C_1 \implies x= \tan (4\log|t+4|+C_2-C_1)$ Similarly, $\arctan x+C_1=4\log|t+4|+C_2 \implies \arctan x+C_1-C_2=4\log |t+4| \implies \frac{1}{4}(\arctan x+C_1-C_2)=\log |t+4|$ Assuming |t+4|>0 we have $\frac{1}{4}(\arctan x+C_1-C_2)=\log |t+4| \implies \exp (\frac{1}{4}(\arctan x+C_1-C_2)) = |t+4|=t+4 \implies \exp (\frac{1}{4}(\arctan x+C_1-C_2))-4=t$2012-05-03
1

Here are some proceeding steps:

  • Seperating the variables you get $\displaystyle \frac{dx}{1+x^2} = 4 \cdot \frac{dt}{t+4}$

  • Integrating both sides you have: $\tan^{-1}(x) + C_{1} = 4 \cdot \log|t+4| + C_{2}$.