# UD-TRG-Maths2-Mumbai-Winter16-Q2-Ans

Qu. 2 (A) Solve \frac{dy}{dx}+\frac yx\log\;y=\frac y{x^2}\left(\log\;x\right)^2

Sol. :- given, \frac{dy}{dx}+\frac yx\log\;y=\frac y{x^2}\left(\log\;x\right)^2

Dividing by y,

\frac1y\frac{dy}{dx}+\frac yx\log\;y=\frac1{x^2}\left(\log\;x\right)^2…..(1)

put v = log y …..(2)

Diff. w.r.x

\therefore\frac{dy}{dx}=\frac1y\frac{dy}{dx}……(3)

Substituting (2) & (3) in equ (1),

\frac{dv}{dx}+\frac1xv=\frac1{x^2}\left(\log\;x\right)^2\;which\;is

of the type \frac{dv}{dx}+pv=\theta

Here, P=\frac1x\;\&\;\theta=\frac1{x^2}\left(\log\;x\right)^2\;

Which are functions of only ‘x’ so given D. equ is linear

Now,

consider, \int P\;dx=\int\frac1xdx=\log\;x

\therefore Integrating\;Factor\;(I.F.)=\int Pdx\;=e^{\log\;x}=x

Therefore General sol. is v\times I.F.=\int\theta\times I.F.\;dx+c

\therefore v.x.=\int\frac1{x^2}\left(\log\;x\right)^2.xdx+c v.x.=\int\frac1x\left(\log\;x\right)^2\times dx+c\;……(A)

put u = log x

\therefore du=\frac1xdx

puting this Value in equ (A)

We get,

\begin{array}{l}v.x.\;=\;\int u^2du\;+\;c\\xv=\frac{u^3}3+c\end{array} \therefore3xv=u^3+3c

putting the value of v & u

\therefore3x\;\log\;y\;=\;\left(\log\;x\right)^3+c

is the General Sol. where \left(c^1=3c\right)

(B) Change the order of integration & evaluate I=\int_0^2\int_\sqrt{2y}^2\frac{x^2dx\;dy}{\sqrt{x^4-4y^2}}

Sol. :- Let, I=\int_0^2\int_\sqrt{2y}^2\frac{x^2dx\;dy}{\sqrt{x^2-4y^2}}

Here, Limits of x are x=\sqrt{2y}\;to\;x=2\; &

Limits of y are y=0\;to\;y=2

x=\sqrt2y\;i.e.\;x^2=2y\;……(1)

is a parabola, while x = 2 ………(2)

is a line parallel to y axis

substituting (2) in (1)

We get,

\begin{array}{l}2^2=2y\\\therefore2\;\;4\;=\;2y\\\therefore Y=2\end{array}

Therefore line x = 2 intersects parabola at A = (2,2)on changing the order of integration, We consider a vertical strip,

Limits of x are y = 0 to y=\frac{x^2}2 &

Limits of x are x = 0 to x = 2

\therefore I=\int_0^2\int_0^{x^2/2}\frac{x^2\;\;dy\;dx}{\sqrt{x^4-4y^2}} =\int_0^2x^2\int_0^{x^2/2}\frac{1\;\;\;dydx}{\sqrt{\left(x^2\right)^2-\left(2y\right)^2}} =\int_0^2x^2\times\frac{\left[\sin\;^{-1}\left({\displaystyle\frac{2y}{x^2}}\right)\right]_0^{x^{2/2}}dx}2 =\frac12\int_0^2x^2\left{\sin\;^{-1}\left(\frac{x^2}2\right)-\sin^{-1}\left(0\right)\right}dx =\frac12\int_0^2x^2\left{\frac{\mathrm\pi}2-0\right}dx =\frac12\times\frac{\mathrm\pi}2\left[\frac{x^3}3\right]_0^2 =\frac12\times\frac{\mathrm\pi}2\left[\frac{\left(2\right)^3-0}3\right] \begin{array}{l}\therefore\frac{\mathrm\pi}{3\times4}\times8\;\;\;2\\I=\frac{2\mathrm\pi}3\\end{array}

Hence, \begin{array}{l}\int_0^2\int_\sqrt{2y}^2\frac{x^2\;dx\;dy}{\sqrt{x^4-4y^2}}=\frac{2\mathrm\pi}3\\end{array}

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