Answer to Question #124212 in Calculus for desmond

Question

(a) Verify that y = e2x sinx is a solution to the differential equation d2y dx2 −4
dy dx
+ 5y = 0.

(b) Differentiate the following functions with respect to x: (i) ln(1 + sin2 x) (ii) xx.

Answer 1

(a) $y=e^{2x} sin x$ is a solution of a differential equation $\frac{d^2y}{ dx^2} −4\frac{dy}{ dx}+ 5y = 0$ if it makes a true equatin when we plug it in.

At first we need to know what the first and the second derivatives are.

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

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

Now we can rewrite the left hand side of our equation by replacing derivatives with their values and simplify the expression:

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

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

$=(8e^{2x} sin x-8e^{2x} sin x)+(4e^{2x} cos x-4e^{2x} cos x)=0.$

We see that the right hand side or our equation is also equals 0.

Since the left hand side and the right hand side are equal, that $y=e^{2x} sin x$ is a solution.

(b)

(i) $ln(1 + sin^2 x)$ is a composite function, so we use the chain rule:

$\frac{d}{dx}[f(g(x))]=f\prime(g(x))g\prime(x).$

$\frac{d}{dx}[ln(1 + sin^2 x)]=\frac{1}{1+sin^2 x}(1+sin^2 x)\prime=\frac{1}{1+sin^2 x}2sin x(sin x)\prime=$

$=\frac{1}{1+sin^2 x}2sin xcos x=\frac{sin2x}{1+sin^2 x}.$

(ii) We'll use a logarithmic differentiation.

$y=x^x,$ hence $ln y=ln(x^x)=x ln x.$

Next we differentiate this expression using the chain rule:

$(ln y)\prime=(x ln x)\prime,$

$\frac{1}{y}y\prime=(x ln x)\prime,$

$y\prime=y(x ln x)\prime=x^x(1ln x+x\frac{1}{x})=x^x(ln x+1).$