Finding function based on slope and point

Question

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?

Answer 1

$$y'(x)=\frac{xy(x)}{1+x^2}\Longleftrightarrow$$ $$\frac{y'(x)}{y(x)}=\frac{x}{1+x^2}\Longleftrightarrow$$ $$\int\frac{y'(x)}{y(x)}\space\text{d}x=\int\frac{x}{1+x^2}\space\text{d}x\Longleftrightarrow$$ $$\ln\left|y(x)\right|=\frac{\ln\left|x^2+1\right|}{2}+\text{C}\Longleftrightarrow$$ $$\left|y(x)\right|=\exp\left[\frac{\ln\left|x^2+1\right|}{2}+\text{C}\right]\Longleftrightarrow$$ $$\left|y(x)\right|=\exp\left[\text{C}\right]\exp\left[\frac{\ln\left|x^2+1\right|}{2}\right]\Longleftrightarrow$$ $$\left|y(x)\right|=\text{C}\exp\left[\frac{\ln\left|x^2+1\right|}{2}\right]\Longleftrightarrow$$ $$\left|y(x)\right|=\text{C}\sqrt{\left|1+x^2\right|}$$

Now, use $y(0)=3$:

$$\left|3\right|=\text{C}\sqrt{\left|1+0^2\right|}\Longleftrightarrow$$ $$3=\text{C}\sqrt{1}\Longleftrightarrow$$ $$3=\text{C}\cdot1\Longleftrightarrow$$ $$3=\text{C}\Longleftrightarrow$$ $$\text{C}=3$$

So:

$$\left|y(x)\right|=3\sqrt{\left|1+x^2\right|}$$

Answer 2

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 $$ 3=C\exp(\ln|\sqrt{1+0}|)=C $$