Find the particular solution for the following differential equation For the following differential equation, find the particular solution:
$$\frac{dy}{dt}=\frac{t-ty}{1+t^2}$$
given: $y(1)=-7 $
I was wondering if someone could help me with the integration? This is how far I have gotten.. 
I tried to solve this using the separation of variables. 
$$\frac{dy}{dt}=\frac{t(1-y)}{1+t^2}$$ then I rewrote it as 
$$\int \frac{t}{1+t^2} dt=\int \frac{1}{1-y} dy $$
then I got $$\frac{1}{2} \ln(1+t^2)=-\ln(1-y)+c $$
I am a little shaky with natural logs so I was wondering if someone could finish it from here for me with a step by step explanation? Thank you!  
 A: Your solution is correct, but its better to use absolute value in logarithm arguments.
If you have trouble to eliminate logarithms,note that your solution can be written as:
$$
\ln\left(\sqrt{1+t^2}\right)=-\ln|1-y|+\ln K \qquad K>0
$$
and if you exponentiate
$$
\sqrt{1+t^2}=\dfrac{K}{1-y}
$$
$$
y=1-\dfrac{K}{\sqrt{1+t^2}}
$$
Since your inital condition is $y(1)=-7$ you can find $K=8\sqrt{2}$

added all steps from the first to the second equation:
$$
\ln\left(\sqrt{1+t^2}\right)=\ln\left(\dfrac{K}{|1-y|}\right) \Rightarrow
$$
$$
\exp\left(\ln \left(\sqrt{1+t^2} \right)\right)=\exp\left(\ln \left(\dfrac{K}{|1-y|}\right)\right) \Rightarrow
$$
$$
\sqrt{1+t^2}=\dfrac{K}{1-y}
$$ 
A: $\text{ using the initial condition is what you mean } \\ \text{b/c you already done the separation of variables part ? } \\  y(1)=-7 \\ \text{ means you are going to replace } \\ t \text{ with } 1 \\ \text{ and } y \text{ with } -7  \text{ and solve for } C. \\ \text{ So you have } \\ \frac{1}{2} \ln|1+1^2|=- \ln|1-(-7)|+C \\ \frac{1}{2} \ln|2|=-\ln|8|+C \\ \text{ not too many steps to follow to find out our constant } \\ \text{ also emilio gave another way to look at the constant }$
