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!
Best Answer
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: $$ \ln\left(\sqrt{1+t^2}\right)=\ln\left(\dfrac{K}{|1-y|}\right) \Rightarrow $$ $$ \exp\left(\ln \left(\sqrt{1+t^2} \right)\right)=\exp\left(\ln \left(\dfrac{K}{|1-y|}\right)\right) \Rightarrow $$ $$ \sqrt{1+t^2}=\dfrac{K}{1-y} $$