<s>[INST] Finding function based on slope and point I have to find the equation of the function given the following information:
$$f(0) = 3 \ \ \ ; \ \ \ \frac{dy}{dx} = \frac{xy}{1+x^2}$$
So:
$$\frac{dy}{y} = \frac{xdx}{1+x^2}$$
Integrate:
$$\ln |y| + K = \ln|\sqrt{1+x^2}| + K$$
$$y = \sqrt{1+x^2} + K$$
Plug in values to find K
$$3 = \sqrt{1+0^2} + K \ \ \ ; \ \ \ K = 2$$
$$y = \sqrt{1+x^2} + 2$$
Answer in the book is 
$$y = 3\sqrt{1+x^2}$$
Where did I go wrong?
 [/INST] You have some confusion in the use of the constant of integration.
1) Wan you integrate $$\frac{dy}{y} = \frac{xdx}{1+x^2}$$ you cannot write the constant $K$ on both sides ($A+K= B+K$ is the same as $A=B$ without $K$), so this step is:
$$\ln |y|  = \ln|\sqrt{1+x^2}| + K$$
2) exponentiating this equation you have:
$$\ y = \exp(\ln|\sqrt{1+x^2}| + K)=\exp(\ln|\sqrt{1+x^2}|)\times \exp(K)=$$
$$
=C\exp(\ln|\sqrt{1+x^2}|)
$$
where $C=\exp(K)$ is a new constant.
3) from the initial condition you can find
$$
3=C\exp(\ln|\sqrt{1+0}|)=C
$$
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