[SOLVED] Integration of Inverse of f(x) [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = [tex]\frac{1}{4y^{3}-18y^{2}+10y}[/tex](ln|[tex]{y^{4}-6y^{3}+5y^{2}}[/tex]|) is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks
[tex]\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] =[tex]\frac{1}{y^{2}(y^{2}-6y+5)}[/tex] =[tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex] So, by Partial fraction, is this correct? [tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex] =[tex]\frac{A}{y}[/tex]+[tex]\frac{B}{y^{2}}[/tex]+[tex]\frac{C}{y-1}[/tex]+[tex]\frac{D}{y-5}[/tex] and then, by comparing coefficient of y^3, y^2, y. (A+C+D){y^{3}}=0 (-6A+B-5C-D){y^{2}}=0 (5A-6B)y=0 answer is A=6/25, B=1/5, C=3/10. D=-87/50. but doubt my answer. anyone can point out?thanks
I can tell you the answer.... no need use latex for such simple expression most likely you have treated the y^2 bit incorrectly
if so, then [tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]=[tex]\frac{6}{25y}[/tex]+[tex]\frac{1}{5y^{2}}[/tex]+[tex]\frac{3}{10(y-1)}[/tex]-[tex]\frac{87}{50(y-5}[/tex]
[tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = 6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right? but this problem is actually make y the subject in this equation, how to go about it further? [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = [tex]\int\[/tex] dt the original question is [tex]\frac{dy}{dt}[/tex]=[tex]y^{4}-6y^{3}+5y^{2}[/tex] find y(t). if [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = 6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y) then i m stuck to make y the subject. anyone to point out the mistake.thanks
another differential eq. question. y'=1+xy find y. how to go about it. i m clueless my method of multiply dx to (1+xy) doesn't work in this case.
You can solve this by the solution for Linear Differential equation. The equation you have is: [tex] \frac{dy}{dx} = 1 + xy [/tex] which can be written as: [tex] \frac{dy}{dx} + y(-x) = 1 [/tex] which is similar to: [tex] \frac{dy}{dx} + yP = Q [/tex] Here, [itex]P = (-x); Q = 1[/itex] So, you have: [tex] I.F = e^{\int (-x)dx} [/tex] [tex] I.F = e^{-\frac{x^2}{2}} [/tex] And hence, your solution is given by: [tex] ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C [/tex] But, it's gonna be a real pain with this: [tex] \int e^{\frac{-x^2}{2}}dx [/tex] a method to which i can't think of right now. Maybe there's some other way..
That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.
great, this is something new to me, just started my linear algebra course.thanks.will study your solution