dy/dx = 1/ (x^2y^2 +xy^2), Thank you :D :D
W warsatan New member Joined Sep 12, 2005 Messages 36 Sep 4, 2006 #1 dy/dx = 1/ (x^2y^2 +xy^2), Thank you
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 4, 2006 #2 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.
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.
W warsatan New member Joined Sep 12, 2005 Messages 36 Sep 4, 2006 #3 Thanks for the quick reply, how do i go about intergration 1/(x^2+x), my brain just went blank.
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,583 Sep 4, 2006 #4 Try partial-fraction decomposition. Eliz.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 5, 2006 #5 Hello, warsatan! How do i go about intergrating: \(\displaystyle \frac{1}{x^2\,+\,x}\) Click to expand... 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: \(\displaystyle \L\,\int\frac{du}{u^2\,-\,a^2}\;=\;\frac{1}{2a}\ln\left|\frac{u\,-\,a}{u\,+\,a}\right|\,+\,C\)
Hello, warsatan! How do i go about intergrating: \(\displaystyle \frac{1}{x^2\,+\,x}\) Click to expand... 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: \(\displaystyle \L\,\int\frac{du}{u^2\,-\,a^2}\;=\;\frac{1}{2a}\ln\left|\frac{u\,-\,a}{u\,+\,a}\right|\,+\,C\)