# Tutor profile: Sahil D.

## Questions

### Subject: Linear Algebra

Let L : P2 ---> P1 be defined by L(p(t)) = p'(t). and consider the ordered bases S = {t^2,t,1} and T = {t,1} for P2 and P1 respectively. (a) Find the matrix A associated with L. (b) If p(t) = 5t ^2 - 3t + 2, compute L(p(t)) directly and then by using A.

(a)We have, L (t^2) = 2t = 2t + 0( 1 ) L (t) = 1 = 0(t) + 1( 1) L (1) = 0 = 0(t) + 0(1) In this case, the coordinates of L(t^2), L(t), and L(1) with respect to the T-basis arc obtained by observation, since the T -basis is quite simple. Thus A= 2 0 0 0 1 0 (b) Since p(t) = 5t^2- 3t + 2, then L(p(t)) = 10t - 3. However, we can find L(p(t)) by using the matrix A as follows: Since 5 [p(t)]s= -3 2 5 then [L(p(t))]T= 2 0 0 -3 = 10 0 1 0 2 -3 which means L(p(t))=10t-3.

### Subject: Calculus

Find the complete solution of differential equation: x^2y’’ -(2m-1)xy’+(m^2+n^2)y=(n^2)(x^m)(lnx) m>0,n>0 and x>0

x^2y’’ -(2m-1)xy’+(m^2+n^2)y=(n^2)(x^m)(lnx)----------------1 Substituting x=e^z in the differential equation 1; d^2y/dz^2-(2m)dy/dz+(m^2+n^2)y=(n^2)(e^mz)(z) For particular solution: Using operator method where in: d/dz=D and d^2/dz^2=D^2 {D^2-(2m)D+(m^2+n^2)}y=(n^2)(e^mz)(z) Where p(D) is polynomial in D p(D)= D^2-(2m)D+(m^2+n^2) y=[1/p(D)](n^2)(e^mz)(z) since n^2 is constant and using exponential shift rule p(D+m)=D^2+n^2 y=(n^2)(e^mz)[1/p(D+m)](z) y=(n^2)(e^mz)[1/ D^2+n^2](z) y=(n^2)(e^mz)[1/(D+in)(D-in)](z) y=(n^2)(e^mz)[1/(D-in)-1/(D+in)](z) y=(n^2)(e^mz)(z/n^2) now back substituting x=e^z y=(x^m)(lnx) now finding general solution: y’’-(2m)y’+(m^2+n^2)y=0 y=c1[(e^mz)(cos(nz))]+c2[(e^mz)(sin(nz))] y=c1[(x^m)cos(nlnx)]+c2[[(x^m)sin(nlnx)] Therefore the solution of the differential equation is: y=c1[(x^m)cos(nlnx)]+c2[(x^m)sin(nlnx)] + (x^m)(lnx) where c1 and c2 are arbitrary.

### Subject: Calculus

Find the domain of function of function: f(x)=ln(ln(lnx)))/(x-3) + sinx

f(x)=ln(ln(lnx)))/(x-3) + sinx Since sinx has a domain of R we need to find domain for ln(ln(lnx)))/(x-3) ln(lnx)>0 => lnx>1 =>x>e And also x-3=/0 Therefore x=/3 The domain of function is (e,infinity)-{3}

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