Cauchy problem: $x'=\frac{x}{t^2+1}, x(0)=1$.

Question

$$x'=\frac{x}{1+t^2}, \qquad x(0)=1$$

I know the solution to the problem, but I don't get to the right solution myself. My solution:

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

$$\ln(x)=\cot(t)+ C$$

$$\ln(1)=\cot(0) + C$$

$$e=C$$

$$\ln(x)=\cot(t) + e.$$

I don't know how to solve the problem further. I think that the constant also is incorrect, because the solution is given as $x=e^{\operatorname{arccot}(t)},\; t \in \mathbb{R}$. Can someone assist with the problem?

Answer 1

You are not far from the solution. But you are being a bit sloppy in the notation, as well as having the wrong antiderivative. First let's rewrite $$\frac{x'}{x}=\frac{1}{1+t^2}.$$ Now, taking the integral over $t$ in both sides gives: $$\int \frac{x'(t)}{x(t)}\,dt=\int \frac{1}{1+t^2}\,dt $$

$$\ln(x(t))=\arctan(t)+c$$

$$x(t)=e^{\arctan(t)+c}$$

And finally, using $x(0)=1$ gives $c=0$.

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As @mathcounterexamples kindly pointed out, one should actually write

$$\ln(|x(t)|)=\arctan(t)+c$$ from which $$|x(t)|=e^{\arctan(t)+c}.$$ and lastly, since $x(0)=1$, $c=0$ and for $x(t)$ to be continuous, $x(t)=e^{\arctan(t)}$

Answer 2

Let:

$$ x = e^f $$

Then:

$$ (e^f)' = \frac{e^f}{1+t^2} $$

$$ f'e^f = \frac{e^f}{1+t^2} $$

$$ f' = \frac{1}{1+t^2} $$

$$ f = arctan(t) +f(0) $$

$$ x = e^{arctan(t)+f(0)} $$

$$ f(0) = ln(x(0)) = ln(1) = 0 $$

Solution:

$$ x = e^{arctan(t)} $$