Question

(a) Find the solution of the initial-value problem and determine its interval of existence (i.e domain of the resulting function y).
(i) dy /dt = 1+t2 /t , y(t-1)=0
(ii) (t + 1)dy /dt = 1−y, y(t = 0) = 3. [Hint: Let A = −(±e−C).]
(b) Evaluate the following integrals: (i) xln(x +1)dx (ii) sin2(ln x)cos2(lnx)/ x dx.

Answer 1

a)(i) Given $\frac{dy}{dt }= \frac{1+t^2}{ t}$

$\implies {dy}= (\frac{1}{ t}+t)dt$

Now by integrating, we get

$y=log(t)+\frac{t^2}{2} + c$

By given condition $y(1)=0$

$\implies c=-\frac{1}{2}$

Hence, solution is $y=log(t)+\frac{t^2}{2} -\frac{1}{2}$ .

log(t) exists for postive values of t, hence interval of existence of solution is $(0,\infin)$ .

a)(ii) Given $(t + 1)\frac{dy}{dt} = 1−y, y(t = 0) = 3$

$\implies \frac{dy}{1-y} =\frac{dt}{t+1}$

By integrating on both sides, we get

$-log(1-y) = log(t+1) -log(c)\\ \implies (1-y)(t+1) =c$

Now, by y(0)=3, $(1-3)(0+1)=c$

$\implies c=-2$ .

Hence solution is $(1-y)(t+1)=-2 \implies y=1+\frac{2}{t+1}$

Solution y is not defined at t=-1 and solution contain y(t=0)=3.

Hence interval of existence of solution $(-1,\infin)$ .

b)(i) Let I = $∫ x ln ( x + 1 ) d x$ = $= \frac{x^2}{2} log ( x + 1 ) − ∫ \frac{x^2}{2}\frac{1}{x+1} d x$

Now for $\int \frac{x^2}{x+1}dx$ , let $u = x+1 \implies dx = du$

So, $\int \frac{x^2}{x+1}dx = \int \frac{(u-1)^2}{u}du$ = $\int (u-2+\frac{1}{u})du = \frac{u^2}{2} -2u+log(u) = \frac{(x+1)^2}{2} - 2(x+1) +log(x+1)$ .

Hence, I = $\frac{x^2}{2}log(x+1)+\frac{(x+1)^2}{4} - (x+1) + \frac{log(x+1)}{2}$

b(ii) Let I = $\int \frac{sin (2(ln x))cos (2(lnx)) }{ x} dx$

Suppose $ln(x) = u \implies \frac{dx}{x} = du$

So, $\int \frac{sin (2(ln x))cos (2(lnx)) }{ x} dx = \int sin (2u)cos (2u) du$

$= \frac{1}{2} \int 2sin (2u)cos (2u) du = \frac{1}{2} \int sin(4u) du$

$= -\frac{1}{8} cos(4u) = -\frac{1}{8} cos(4 \ ln(x))$