Answer to Question #124029 in Calculus for Nestor Freeman

Question

(a) Verify that y = e2xsinx is a solution to the differential equation d2y/dx2 - 4dy/dx+5y=0

(b) Differentiate the following functions with respect to x: (i) ln(1 + sin2 x) (ii) xx.
(c) Evaluate the integral 2x3 −4x−8 /x4 −x3 +4x2 −4x dx.

Answer 1

(a) $y(x)=e^{2x} sin x.$

Let's differentiate this function.

$\frac{dy}{dx}=(e^{2x})^\prime sin x+e^{2x}(sin x)^\prime=2e^{2x}sin x+e^{2x}cos x.$

$\frac{d^2y}{dx^2}=2(e^{2x})^\prime sin x+2e^{2x}(sin x)^\prime+(e^{2x})^\prime cos x+ e^{2x}(cos x)^\prime=$

$=4e^{2x} sin x+2e^{2x}cos x+ 2e^{2x}cos x-e^{2x} sin x=3e^{2x} sin x+4e^{2x}cos x.$

Hence we have:

$\frac{d^2y}{dx^2} - 4\frac{dy}{dx}+5y=3e^{2x} sin x+4e^{2x}cos x-4(2e^{2x}sin x+e^{2x}cos x)+5e^{2x}sin x=$

$=3e^{2x} sin x+4e^{2x}cos x-8e^{2x}sin x-4e^{2x}cos x+5e^{2x}sin x=0,$

so that the function $y(x)=e^{2x} sin x$ is a solution to the differential equation $\frac{d^2y}{dx^2} - 4\frac{dy}{dx}+5y=0.$

(b)

• (i) $(ln(1 + sin^2x))^\prime =\frac{1}{1+sin^2x}(1+sin^2x)^\prime=$

$\frac{1}{1+sin^2x}2sin xcos x=\frac{sin 2x}{1+sin^2x}.$

• (ii) $(x^x)^\prime=(e^{x ln x})^\prime=e^{x ln x}(x ln x)^\prime=e^{x ln x}((x)^\prime ln x+ x (ln x)^\prime)=$

$=e^{x ln x}(1ln x+x\frac{1}{x})=e^{x ln x}(ln x+1)=x^x (lnx+1).$

(c) We'll use the method of partial fractions to evaluate the integral $\int \frac{2x^3 −4x−8}{x^4 −x^3 +4x^2 −4x} dx.$

$\frac{2x^3 −4x−8}{x^4 −x^3 +4x^2 −4x}=\frac{2x^3 −4x−8}{x^3(x-1)+4x(x-1)}= \frac{2x^3 −4x−8}{(x^3+4x)(x-1)}=$

$=\frac{2x^3 −4x−8}{x(x-1)(x^2+4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+4}.$

Then we have:

$A(x-1)(x^2+4)+Bx(x^2+4)+(Cx+D)x(x-1)=2x^3-4x-8.$

$A(x^3+4x-x^2-4)+B(x^3+4x)+C(x^3-x^2)+D(x^2-x)=2x^3-4x-8.$

$(A+B+C)x^3+(-A-C+D)x^2+(4A+4B-D)x+(-4A)=2x^3-4x-8.$

Hence:

$A+B+C=2,$

$−A−C+D=0,$

$4A+4B−D=−4,$

$−4A=−8.$

From the last equation: $A=2$. We have:

$B+C=0, D-C=2, 4B-D=-12.$

Then $B=-C, D=C+2,$

and we have the next equation: $-4C-2-C=-12,$

$-5C=-10, C=2.$

Then $B=-C=-2, D=C+2=2+2=4.$

So we have:

$\frac{2x^3 −4x−8}{x^4 −x^3 +4x^2 −4x}=\frac{2}{x}-\frac{2}{x-1}+\frac{2x+4}{x^2+4}.$

Now we can evaluate the integral:

$\int \frac{2x^3 −4x−8}{x^4 −x^3 +4x^2 −4x} dx=\int(\frac{2}{x}-\frac{2}{x-1}+\frac{2x+4}{x^2+4})dx=$

$=\int\frac{2}{x}dx-\int\frac{2}{x-1}dx+\int\frac{2x+4}{x^2+4}dx=\int\frac{2}{x}dx-\int\frac{2}{x-1}dx+\int\frac{2x}{x^2+4}dx+\int\frac{4}{x^2+4}dx=$

$=\int\frac{2}{x}dx-\int\frac{2}{x-1}dx+\int\frac{d(x^2+4)}{x^2+4}+\int\frac{4}{x^2+4}dx=$

$=2ln|x|-2ln|x-1|+ln|x^2+4|+4\cdot \frac{1}{2}arctan\frac{x}{2}+C=$

$=2ln|x|-2ln|x-1|+ln|x^2+4|+2arctan\frac{x}{2}+C.$