Answer to Question #124024 in Calculus for Joseph Ocran

Question

Verify that y=e^2xsinx is a solution to the differential equation (d^2/dx^2)-(4dy/dx)+5y=0


(2) Differentiate ln(1+sin^2x)

(3) differentiate x^x
(4).Evaluate the integral [ (2x^3-4x-8)/(x^4-x^3+4x^2-4x)]dx

Answer 1

Given $y=e^{2x}sinx \implies y' = e^{2x}cosx+2e^{2x}sinx$

$\implies y''-4y'+5y = e^{2x}(4cosx+3sinx-4cosx-8sinx+5sinx) = e^{2x} (0) = 0$

Hence, $y =e^{2x} sinx$ is solution of $y''-4y'+5y=0$ .

2) Let $y = ln(1+sin^2x)$

So, by differentiation with respect to x, we get

$\frac{dy}{dx} = \frac{1}{1+sin^2x} (2sinxcosx)$ .

3) Let $y = x^x$ $\implies log(y)= xlog(x)$

So, by differentiation with respect to x, we get

$\frac{1}{y}\frac{dy}{dx} = 1+log(x) \implies \frac{dy}{dx} = x^x(1+logx)$ .

4) We partial fraction,

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

Hence, $\intop\frac{\left(2x^3-4x-8\right)}{\left(x^4-x^3+4x^2-4x\right)} dx= \intop( \frac{2}{x}-\frac{2}{x-1}+\frac{2x}{x^2+4} + \frac{4}{x^2+4})dx$

= $2ln(x)-2ln(x-1)+ln(x^2+4)+ 2tan^{-1}(\frac{x}{2}) +c=$

$= ln(\frac{x^2(x^2+4)}{(x-1)^2}) +2tan^{-1}(\frac{x}{2})+c$ .