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UD-TRG-Maths2-Mumbai-Winter16-Q5-Ans

Qu. 5 (A) Use method of variation of parameters to solve

\frac{d^2y}{dx_2}-5\frac{dy}{dx}+6y\;=\;e^{2x}x^2.

Sol. :- given, \frac{d^2y}{dx_2}-5\frac{dy}{dx}+6y\;=\;e^{2x}x^2.

It’s Auxiliary equ is D^2-5D+6=0

put D = m

m^2-5m+6=0

on solving

m = 2 or m = 3

\begin{array}{l}\therefore Complimeutory\;F^u\;\left(c.f.\right)\;is\\yc=c_1e^{2x}+c_2e^{3x}\end{array}

let,

\begin{array}{l}y_1=e^{2x},\;y_2=e^{3x}\\x\;=\;e^{2x}\;x^2\end{array} \begin{array}{l}\therefore y_1^1=e^{2x}\times2\;=\;2e^{2x}\;\\\&\;y_2^1=e^{3x}\times3\;=\;3e^{3x}\end{array}

Now,

W=\begin{vmatrix}y_1&y_2\y_1^1&y_2^1\end{vmatrix} \begin{array}{l}=y_1\;y_{2\;}^1-\;y_1^1\;y_2\\=e^{2x}\times3e^{3x}-2e^{2x}\times e^{3x}\\=3e^{5x}-2e^{5x}\end{array} W=e^{5x} \begin{array}{l}u\;=\;\int\frac{y_2x}w\;dx\\=\int\frac{e^{3x}}{e^{5x}}\;e^{2x}\times x^2\;dx\end{array} \begin{array}{l}\;=\;\int x^2\;dx\\=\frac{x^3}3\end{array}

& \begin{array}{l}V\;=\;\int\frac{y_1x}w\;dx\\=\int\frac{e^{2x}}{e^{5x}}\times e^{2x}\;x^2\;dx\end{array}

\begin{array}{l}\;=\int x^2\times e^{-x}\;dx\\V\;=\;x^2\times\frac{e^{-x}}{-1}-2x\times\frac{e^{-x}}{\left(-1\right)^2}+2\frac{e^{-x}}{\left(-1\right)^3}\end{array} \begin{array}{l}\;=-x^2e^{-x}-\;2xe^{-x}-2e^{-x}\\V=\;-e^{-x}\left[x^2+2x+2\right]\end{array}

particular integral (P.I.) yp=uy_1+vy_2

\therefore yp=-\frac{x^3}3\times e^{2x}-e^{-x}\left[x^2+2x+2\right]\times e^{3x} =-\frac{x^3}3e^{2x}-\left[x^2+2x+2\right]\times e^{2x} yp=\frac{-1}3e^{2x}\left[x^3+3\left(x^2+2x+2\right)\right] \begin{array}{l}\therefore The\;complete\;sol.\;is\;y\;=\;yc\;+\;yp\\Y=c_1e^{2x}+c_2e^{3x}-\frac13e^{2x}\left[x^3+3x^2+6x+6\right]\end{array}

(B) Using Taylor’s series method, obtain the sol. of

\frac{dy}{dx}=3x+y^2,\;y(0)=1 find the value of y for x = 0.1 correct to

four decimal places.

Sol. :- let, \begin{array}{l}y^1=\frac{dy}{dx}=3x+y^2\\y^{11}=3+2y\;y^1\end{array}

\begin{array}{l}y^{111}=2\left(yy^{11}+y^1y^1\right)\\=2\left(yy^{11}+\left(y^1\right)^2\right)\end{array} \begin{array}{l}y^{iv}=2\left(yy^{iii}+y^{ii}\;y^i+2y^i\;y^{ii}\right)\\=2\left(yy^{iii}+3y^i\;y^{ii}\right)\end{array} \begin{array}{l}y^v=2\left(yy^{iv}+y^{iii}y^i+3y^iy^{iii}+3y^{ii}y^{ii}\right)\\=2\left(yy^{iv}+4y^iy^{iii}+\left(3y^{ii}\right)^2\right)\end{array}

Now,

\begin{array}{l}At\;\;x0\;=\;0\;,\;y0\;=\;1\\y0^1=3x_0+y_0^2\\=3\left(0\right)+\left(1\right)^2\\end{array} \begin{array}{l}y0^i=1\\y0^{ii}=3+2y0y0^i\\=3+2\;\left(1\right)\left(1\right)\\y0^{ii}=5\\end{array} \begin{array}{l}y0^{iii}=2\left(y0\;y0^{ii}+y_0^i^2\right)\\=2\left(1\times5+\left(1\right)^2\right)\\y0^{iii}=12\end{array} \begin{array}{l}y0^{iv}=2\left(y0\;y0^{iii}+3y0^iy0^{ii}\right)\\=2\left(1\times12+3\times1\times5\right)\\y0^{iv}=54\end{array} \begin{array}{l}y0^v=2\left(y0\;y0^{iv}+4y0^iy0^{iii}+\left(3y^{ii}\right)^2\right)\\=2\left[1\times54+4\times1\times12+3\times\left(5\right)^2\right]\\y0^v=354\end{array}

Therefore By Taylor series method,

Y=y0+xy_0^i+\frac{x^2}{21}y_0^{ii}+\frac{x^3}{31}y_0^{iii}+\frac{x^4}{41}y_0^v+….. \therefore Y=1+x\left(1\right)+\frac{x^2}{21}\left(5\right)+\frac{x^3}{31}\left(12\right)+\frac{x^4}{41}\left(54\right)+\frac{x^5}{51}\left(354\right) \therefore Y=1+x+\frac52x^2+2x^3+\frac94x^4+\frac{59}{20}x^5+….

(C) Find the value of integral \int_0^1\frac{x^2}{1+x^3}\;dx\;dy by taking h = 0.2, using (1) Trapezoidal rule (2) Simpson’s \frac13 Rule. Compare the errors with th exact value of the integral.

Sol. :- let, a = 0, b = 1, & h = 0.2

let, y=\;\frac{x^2}{1+x^3}

\begin{array}{l}\therefore h=\frac{b-a}h\;i.e.\;\;n=\frac{b-a}h\\\therefore\;n=\frac{1-0}{0.2}\\n=5\end{array}
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