icedpolonium
New member
- Joined
- Sep 10, 2015
- Messages
- 1
Hello 
I'm having difficulties with the following differential equation:
\(\displaystyle y'={e^{x^2}}\log(y-2)\)
I tried to solve it by separating the variables (\(\displaystyle y>2, y \neq 3\)):
\(\displaystyle \frac{\partial y}{\partial x}={e^{x^2}}\log(y-2)\)
\(\displaystyle \frac{\partial y}{\partial x}\frac{1}{\log(y-2)}={e^{x^2}}\)
and then I'm stuck after trying to integrate both sides with respect to x:
\(\displaystyle \int \frac{\partial y}{\log(y-2)}=\int {e^{x^2}} \partial x\)
In fact, I can't integrate even one of them
If it is of any help, the ODE was part a Cauchy problem, with \(\displaystyle y(0)=3\) being the condition to satisfy.
Am I missing something?
I'm having difficulties with the following differential equation:
\(\displaystyle y'={e^{x^2}}\log(y-2)\)
I tried to solve it by separating the variables (\(\displaystyle y>2, y \neq 3\)):
\(\displaystyle \frac{\partial y}{\partial x}={e^{x^2}}\log(y-2)\)
\(\displaystyle \frac{\partial y}{\partial x}\frac{1}{\log(y-2)}={e^{x^2}}\)
and then I'm stuck after trying to integrate both sides with respect to x:
\(\displaystyle \int \frac{\partial y}{\log(y-2)}=\int {e^{x^2}} \partial x\)
In fact, I can't integrate even one of them
If it is of any help, the ODE was part a Cauchy problem, with \(\displaystyle y(0)=3\) being the condition to satisfy.
Am I missing something?