Homework Statement
If y(1+x2) dy/dx = 2x (1-y2), prove that (1+x2)2(1-y2)=A, where A is constant.
Homework Equations
Separable equations
The Attempt at a Solution
Separate the terms:
y/(1-y2) dy = 2x/(1+x2) dx
Integrating both sides will get:
∫ y/(1-y2) dy = ∫...
So, substituting x = 2, A = 1 and C = 1, I will get:
1 = A(2-1)^2 + B(2)(2-1) + C(2)
1 = A^2 + 2B + 2C
1= 1^1 + 2B + 2
2B = -2
B = -1
Got it, thanks a lot. Much appreciated. Learned something new here. When you say "can use any other value to get an equation for B", can it be implemented...
oh dayum...dayum...d.a.y.u.m...
Thank Kreizhn for pointing that out, ouch.
So from the method that you were saying:
1= A(x-1)2 + Bx(x-1) + Cx
Substitute x=0 to the equation, I will get:
1 = A(0-1)2 + B(0)(0-1) + C(0)
1= A(-1)^2
A=1
Substitute x =1 to the equation, I will...
All right, thanks guys. I've also done the substitution part.
u=x-1
du=dx
Finally got it.
At Kreizhn: Its "1 = A(x-1)^2 + Bx(x-1) + Cx", not "1 = A(x-1)2 + Bx(x-1) + Cx" or maybe you meant something else? :P
ok, continuin with the given information from the people above:
1 = A(x2-2x+1) + Bx2-B + Cx
1 = Ax2-2Ax+A + Bx2-B + Cx
Sorting through, then I will get:
1 = Ax2 + Bx2 -2Ax + Cx + A - B
1 = (A+B)x2 - (2A-C)x + (A-B)
From the above equation:
A + B = 0
2A - C = 0
A - B =...
Homework Statement
Evaluate ∫1/x(x-1)2, by using the partial fraction method2. The attempt at a solution
Stating in partial fraction form:
1/(x(x-1)2 = A/x + B/(x-1) + C/(x-1)2
1 = Ax(x-1)2/x + Bx(x-1)2/(x-1) + Cx(x-1)2/(x-1)2
1 = A(x-1)2 + Bx(x-1) + Cx
and this is where I am stuck...