kalyani –

Q.1 (a) show that \int_0^\infty3^{-4x^2}\operatorname dx=\frac{\sqrt{\mathrm\pi}}{4\;\log3}

Solution:- Let I = \int_0^\infty3^{-4x^2}\operatorname dx

put 3^{-4x^2} = e^-v _________ A

Taking log on both side

\log3^{4x^2}=\log e^{-v}\\

-4x² log3 = -V loge _______________1

x^2=\frac v{4\log3}\;\;\;\;\cdots\cdots\cdots\;\left(\log e=1\right)\\\\\\

Taking Square root

x = \frac{V^{\displaystyle\frac12}}{2\sqrt{\log3}}

dx =\frac1{2\;\sqrt{\log\;3}}\times\frac12V^{-\frac12dv}

from (1) ____________(B)

V = 4x² log3

Taking new limits

When x=0, V=0,

x=\infty,\;V=\infty _____ from(A) & (B)

I= \int_0^\infty e^{-v}\times\frac1{2\sqrt{\log3}}\times\frac12v^\frac12dv

= \frac1{4\sqrt{\log3}}\int_0^\infty e^{-v}v^{-\frac12}dv

=\frac1{4\sqrt{\log3}}\times\sqrt{\frac12}

\int_0^\infty e^{-x}x^4\operatorname dx=\sqrt{x+1} \sqrt{\frac12}=\sqrt{\mathrm\pi}

I= \frac1{4\log3}\times\sqrt{\mathrm\pi}

Hence, \int_0^\infty3^{-4x}dx=\frac{\sqrt{\mathrm\pi}}{4\;\log3}

Q.1 (b) Solve (2y² -4x+5)dx =(y-2y² -4xy)dy

Solution:- let,

M=2y²-4x+5

Diff. M w.r.t y

\frac{\operatorname dM}{\operatorname dy}=4y-0+0\\\\\ \frac{\operatorname dM}{\operatorname dy}=4y\;\;\;\;\;\;\cdots\cdots\cdots\cdots\cdots\left(A\right)\\\\\

N= y-2y²-4xy

Diff. N w.r.t x

\frac{\operatorname dN}{\operatorname dx}=0-0-4y\;\;\;\;\;\;\;\;\;\cdots\cdots\cdots\cdots\cdots\cdots\cdots\left(B\right)\\\\\

from equⁿ (A)&(B)

given Differential equⁿ is exact

Genenral solution is

\int_{y=cons\tan t}M\operatorname dx\;+\;\int\left(term\;of\;N\;free\;=c\;from\;x\right)\;dy\\\\\\ \int_{y=cons\tan t}\left(2y^2-4x+5\right)dx\;+\int\left(y-2y^2\right)dy\;=c\\\\\\ -4\int xdx\int2y^2\int dx+5\int dx+\int ydy\;-2\int y^{2dy}\\\\\\ -4^2\frac{x^2}2+2y^2x+\;5x+\frac{y^2}2-\frac{2y^3}3=c\\\\\

-2x²+2xy²+5x+\frac{y^2}2\;-\frac{2y^2}3=c

12x^2-12xy^2-30x-\frac{6y^2}2+\frac{12y^3}3=-6c

12x²-12xy²-30x-3y²+4y³=-6c

12x²-30x-3y²-12xy²=c

is the general solⁿ

d) Evaluate \int_0^1\int_0^{x^2}e^\frac yx\operatorname dydx\;=\frac12

Solⁿ :- Let I=\int_0^1\int_0^1e^\frac yx\operatorname dydx

Integrate with respect to y

=\int_0^1\left[\frac{e{\displaystyle\frac yx}}{\displaystyle\frac1x}\right]^{x^2}dx

=\int_0^1\frac1{\displaystyle\frac1x}\left[e^{x\frac1x}-e^0\right]dx

= \int_0^1x\left[e^x-1\right]dx

=\int_0^1\left[xe^x-x\right]dx

Integrate with respect to x

={\left[\left(xe^x-1.e^x\right)-\frac{x^2}2\right]}_0^1

= \left[\left(1.e^1-e^1\right)-\frac12\right]-\left[\left(0-e^0\right)-0\right]

= \frac{-1}2+1

I=\frac12

Hence proved \int_0^1\int_0^{x^2}e^\frac yx\operatorname dydx\;=\frac12

Que3. a) Evaluate \int_0^1\int_0^{1-x}\int_0^{1-x-y}\frac1{\left(x+y+x+1\right)^3}dzdydx

Solⁿ :- Let I=\int_0^1\int_0^{1-x}\int_0^{1-x-y}\frac1{\left(x+y+z+1\right)^3}dzdydx\\

I=\int_0^1\int_0^{1-x}\int_0^{1-x-y}\left(x+y+z+1\right)^{-3}dzdydx\\

Integrating w.r.t z

We get

I=\int_0^1\int_0^{1-x}{\left[\frac{\left(x+y+z+1\right)}{-2}\right]}_0^{1-x-y}dydx\\ I=\int_0^1\int_0^{1-x}\frac1{-2}\left[\left(x+y+1-x-y+1\right)^{-2}-\left(x+y+0+1\right)^{-2}\right]dydx\\ I=-\frac12\int_0^1\int_0^{1-x}\left[2^{-2}-\left(x+y1\right)^{-2}\right]dydx\\

Integrating w.r.t y

I=-\frac12\int_0^1{\left[\frac14y-\frac{\left(x+y+1\right)^{-1}}{-1}\right]}_0^{1-x}dx\\

Que5.c) Compute the value of \int_{0.2}^{1.4}\left(\sin x-lux\;+e^x\right)dx using 1) Trapezoidal Rule 2) Simpson’s \left(\frac13\right)^{rd} 3) Simpson’s \left(\frac38\right)^{th} rule by dividing into six subintervals.

Solⁿ :- Let a=0.2, b=1.4 & x=6

Let y=sinx-lux+e^x

h=\frac{b-a}x=\frac{1.4-0.2}6=\;0.2

1)By using Trapezoidal Rule:-

\int_a^bf\left(x\right)dx =\frac h2\left[\left(y_0+y_6\right)+\left(y_1+y_2+y_3+y_4+y_5\right)\right]

=\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx

=\frac{0.2}2\left[\left(3.0295+4.7042\right)+2\left(2.7975+2.8976+3.1660++3.5598+4.0698+\right)\right]

=0.1\left(7.7337+2\times21.1949\right)

\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx\;=4.0715

2) By using Simpson,s \left(\frac13\right)^{rd} Rule :-

\int_a^bf\left(x\right)dx =\frac h3\left[\left(y_0+y_6\right)+4\left(y_1+y_3+y_5\right)+2\left(y_2+y_4\right)\right] \therefore\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx \frac{0.2}3\left[\left(3.0295+4.7042\right)+4\left(2.7975+3.1660+4.0698\right)+2\left(2.8976+3.5598\right)\right]

=\frac{0.2}3\left[7.7337+4\times10.0333+2\times6.4574\right]

\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx=4.0521

3)By using Simpson’s \left(\frac38\right)^{th} Rule:-

\int_a^bf\left(x\right)dx \left[\left(y_0+y_6\right)+3\left(y_1+y_2+y_4+y_5\right)+2y_3\right] \therefore\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx

=\frac{3\times0.2}8\left[\left(3.0295+4.7042\right)+3\left(2.7975+2.8976+3.5598+4.0698\right)+2\left(3.1660\right)\right]

=\frac{0.6}8\left[7.7337+3\times13.3247+2\times3.166\right] \;\;\boxed{\;\;\int_{0.2}^{1.4}\left(\sin x-lux+e^x\right)dx\;\;=4.0530\;\;\;}

Q.1 (c) Solve the ODE (D-1)² (D+1)²y=0

Solⁿ:- The given D. equⁿ is

(D-1)² (D+1)²y=0 ________(1)

The Auxiliary equⁿ is

put D=m _______in(1)

(m-1)² =0 or (m²+1)=0

m=1 m²=-1=i²

m=1 m=\pm i