# Differential equation initial value problem - hard!!

I have been asked to solve $x' = t/(1 + t^2) - x(t/(1+t^2))$ and determine the maximal interval where the solution exists.

I have tried to solve this in many different ways but must be using the wrong method, could someone please explain which way i would use to solve this as im pretty sure i am getting it completely wrong!! I know how to solve differential equations normally, so shouldnt need too much of an explanation... just which method to use to solve! I think it is a linear equation but i may be wrong.

it is due it at 4pm so need some help asap!! Thanks in advance :)

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I assume the last part is $x$ times $t/(1+t^2)$ and not the value of $x$ at that point. Then the equation is linear. You need to look for an integrating factor. – Harald Hanche-Olsen Feb 19 '13 at 14:00
yes it is x times t/(1+t2), i have tried intergrating factor and i dont think it works.... if you are sure that is correct can you please explain? thanks - Im assuming it is linear as f(t,x) = a(t)x + b(t) ?? – camilla Feb 19 '13 at 14:02

The homogeneous equation is $x'=-x\cdot t/(1+t^2)$. Handled as a separable equation, that yields the solution $$\ln x=-\int\frac{t}{1+t^2}\,dt.$$ So the integrating factor should be $e^{\phi(t)}$, where $\phi(t)$ is the integral above (without the minus sign, and you can drop any constant of integration of course).

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Hint: Multiply both sides of the equation by $$s^\prime(t)=\left( \frac{t}{1 + t^2}\right)^\prime.$$ Make the change of parameters putting 't' as a function of $s(t)=\frac{t}{1 + t^2}$ and then see coresponding equation and its solution in EqWord.

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The integrating factor of the equation is $\sqrt{1+t^2}$. This may be found by rearranging the equation as

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

The integrating factor is given by $e$ to the integral of the coefficient of $x$:

$$\exp{\left [\int dt \frac{t}{1+t^2} \right ]} = \exp{\left [\frac{1}{2} \log{(1+t^2)}\right] }= \sqrt{1+t^2}$$

Multiplying the original equation by this factor produces

$$\frac{d}{dx} [x \sqrt{1+t^2}] = \frac{t}{\sqrt{1+t^2}}$$

Integrate both sides and do not forget a constant of integration.

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