Find general solution for this separable equation.
- Thread starter warsatan
- Start date
Start by factoring in the denominator.
\(\displaystyle \L\\\frac{dy}{dx}=\frac{1}{x^{2}y^{2}+xy^{2}}\)
\(\displaystyle \L\\\frac{dy}{dx}=\frac{1}{y^{2}(x^{2}+x)}\)
Now, separate variables, integrate and all that good stuff.
\(\displaystyle \L\\\frac{dy}{dx}=\frac{1}{x^{2}y^{2}+xy^{2}}\)
\(\displaystyle \L\\\frac{dy}{dx}=\frac{1}{y^{2}(x^{2}+x)}\)
Now, separate variables, integrate and all that good stuff.
Hello, warsatan!
Eliz. Stapel had the best approach . . . Here's another:
Complete the square: \(\displaystyle \,x^2\,+\,x\;=\;x^2\,+\,x\,+\,\frac{1}{4}\,-\,\frac{1}{4} \;=\;\left(x\,-\,\frac{1}{2}\right)^2 \,-\,\frac{1}{4}\)
The integral becomes: \(\displaystyle \L\,\int\frac{dx}{\left(x\,-\,\frac{1}{2}\right)^2\,-\,\left(\frac{1}{2}\right)^2}\)
Then use the formula: \(\displaystyle \L\,\int\frac{du}{u^2\,-\,a^2}\;=\;\frac{1}{2a}\ln\left|\frac{u\,-\,a}{u\,+\,a}\right|\,+\,C\)
Eliz. Stapel had the best approach . . . Here's another:
Complete the square: \(\displaystyle \,x^2\,+\,x\;=\;x^2\,+\,x\,+\,\frac{1}{4}\,-\,\frac{1}{4} \;=\;\left(x\,-\,\frac{1}{2}\right)^2 \,-\,\frac{1}{4}\)
The integral becomes: \(\displaystyle \L\,\int\frac{dx}{\left(x\,-\,\frac{1}{2}\right)^2\,-\,\left(\frac{1}{2}\right)^2}\)
Then use the formula: \(\displaystyle \L\,\int\frac{du}{u^2\,-\,a^2}\;=\;\frac{1}{2a}\ln\left|\frac{u\,-\,a}{u\,+\,a}\right|\,+\,C\)