Answer to Question #218090 in Calculus for Mohammed Ariful Is

Question

1) Find the following limit:  lim (x→0) ln(1 + (sin(2x))^2 )/ (1 − cos2(x)).

2) Determine type of the differential equation y`` − 2y` + y = sin x:

◦ partial differential equation.

◦ first order differential equation.

◦ linear differential equation with constant coefficients.

◦ linear nonhomogeneous differential equation.

◦ nonlinear homogeneous differential equation.

3) Write general solution of the differential equation x^(2) y'' + xy' + a^(2) y = 0:

◦ Ax^2 + Bx

◦ Ax^a + Bx^−a

◦ Ax^ia + Bx^−ia

◦ Ae^ax + Be^−ax

◦ Ae^iax + Be^−iax

◦ explicit algebraic form does not exist.

Answer 1

1)

$\lim\limits_{x\to 0}\dfrac{\ln(1+\sin(2x))^2}{1-\cos^2(x)}=\lim\limits_{x\to 0}\dfrac{2\ln(1+\sin(2x))}{\sin^2(x)}$

$\lim\limits_{x\to 0}(2\ln(1+\sin(2x))=0$

$\lim\limits_{x\to 0}(\sin^2(x))=0$

$\dfrac{0}{0}$

$\lim\limits_{x\to 0}\dfrac{\ln(1+\sin(2x))^2}{1-\cos^2(x)}=\lim\limits_{x\to 0}\dfrac{2\ln(1+\sin(2x))}{\sin^2(x)}$

$=2\lim\limits_{x\to 0}\dfrac{(\ln(1+\sin(2x)))'}{(\sin^2(x))'}$

$=2\lim\limits_{x\to 0}\dfrac{\dfrac{2\cos(2x)}{1+\sin(2x)}}{2\sin(x)\cos(x)}$

$=2\lim\limits_{x\to 0}\dfrac{\cos(2x)}{\sin(x)\cos(x)(1+\sin(2x))}$

$\lim\limits_{x\to 0^-}\dfrac{\ln(1+\sin(2x))^2}{1-\cos^2(x)}=-\infin$

$\lim\limits_{x\to 0^+}\dfrac{\ln(1+\sin(2x))^2}{1-\cos^2(x)}=+\infin$

2)

$y''− 2y' + y = \sin x$

This is linear nonhomogeneous differential equation.

3)

$x^2 y'' + xy'+ a^2 y = 0$

Let $a=0$

$x^2 y'' + xy'= 0$

$y'=z, y''=z'$

$x^2 z'+ xz= 0$

$\dfrac{dz}{z}=-\dfrac{dx}{x}$

$\ln|z|=-\ln|x|+\ln C_1$

$z=\dfrac{C_1}{x}$

$y'=\dfrac{C_1}{x}$

$y=C_1\ln x+C_2$

Therefore the answer is

explicit algebraic form does not exist.