Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$

Question

Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$

I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$

$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$

$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$

$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$

From here, I got stuck. I have to remove y from here to solve it.

The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?

Answer 1

solving the equation $$\ln(1+2y)=ln(1+x^2)+C$$ for $y$ we get $$y=\frac{e^c}{2}(x^2+1)-\frac{1}{2}$$ now substitute $$k=\frac{e^C}{2}$$

Answer 2

First double both sides to delete the halves, then exponentiate. Thus $k=\pm\exp 2C$, the sign being due to removing the modulus signs.

Answer 3

To get rid of the ln on both sides just do $e^{ln|1+2y|}$

Answer 4

$$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$

$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + \frac {1}{2} \ln C$$

$$ | 1 + 2y | =C(1+x^2)$$

$$ 2y = -1 +C(1+x^2)$$

$$ y = \frac {-1 +C(1+x^2)}{2}$$

Answer 5

You got : $\quad\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $ $$\frac{1}{2} \ln | 1 + 2y | - \frac{1}{2} \ln | 1+x^2 | = C $$ $$\ln | 1 + 2y | - \ln | 1+x^2 | = 2C $$ $$\ln \frac{| 1 + 2y |}{ | 1+x^2 |} = 2C $$ Thus $\quad\frac{ 1 + 2y }{ 1+x^2 }=$constant. $$\frac{ 1 + 2y }{ 1+x^2 }=c$$ $$1+2y=c(1+x^2)$$ $$y=\frac12\left(c(1+x^2)-1\right)$$