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.