Homework SourseFind the partial fraction decompostion 59x225Solution5x2 25
Find the partial fraction decompostion 5/9x2-25
Solution
5/(x^2 - 25) = 5/(x-5)(x+5)
= A/(x+5) +B/(x-5)
= [A(x-5) +B(x+5)]/(x^2 -25)
equating the coefficients on both sides
5= x(A +B) -5A +5B
A+B =0\'
-5A +5B =1----> -A +B =1
we get values of A and B : B = 1/2 ; A = -1/2
5/(x^2 - 25) = -1/2(x+5) +1/2(x-5) ( partial fractions)
![Find the partial fraction decompostion 5/9x2-25Solution5/(x^2 - 25) = 5/(x-5)(x+5) = A/(x+5) +B/(x-5) = [A(x-5) +B(x+5)]/(x^2 -25) equating the coefficients on](/WebImages/31/find-the-partial-fraction-decompostion-59x225solution5x2-25-1087869-1761572313-0.webp "Find the partial fraction decompostion 5/9x2-25Solution5/(x^2 - 25) = 5/(x-5)(x+5) = A/(x+5) +B/(x-5) = [A(x-5) +B(x+5)]/(x^2 -25) equating the coefficients on")
