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Solve (D2-3D+2)y =x(x+4) and show that its general solution is given by y=Aex + Be2x +(x2/2) + (7x/2) + (19/4)
My work :
auxiliary eqn , m2 - 3m + 2= 0 m=1 , m=2
Reduced eqn : yn = Aex + Be2x
P.I. = 1/f(D)F(x) = (1/(D2-3D+2))(x(x+4))
P.I. = [(x2+4x) + ( (1/2)( 2- 3(2x) - 3(4) ) ) + (1/4)(92) )
P.I. = x2 + (4x/2) - (6x/4) - (5/2) + (18/4)
P.I. =(x2/2) + x + 8/4
General solution : y= Aex + Be2x + (x2/2) + x + 8/4
Could anyone point out my mistakes?
It is given that a,b,c are positive values. 1) show that (a+b+c)((1/a)+(1/b)+(1/c)) >= 9
2) If a+b+c=1 , show that (2-a),(2-b),(2-c) are positive
3) Using above results show that [(a/(2-a)) + (b/(2-b)) + (c/(2-c)) ] >= 3/5
I need help with 3rd part
Integrate [x4(1+x)4/(1+x2) ] and explain why π≠22/7?
After integrating I got( π+ rational number)(let us take this as "t".But how do we find a value(let us take this as I) for the integral? Can we get a inequality by using the given limit such as; when x--->0 , I----->0 and when x---->1 , I---->[ 1*(1+1)4]/(1+1)=8 and take as, 0