UD-TRG-Maths2-Mumbai-Winter16-Q4-Ans
Qu. 4 (A) Show that the length of the arc of the parabola y2=4ax cut off by the line 3y=8x\;is\;a\;\left[\log^2+\frac{15}{16}\right]
Sol. :- y^2=4ax\;……(1)\;\&
line 3y = 8x
\therefore x=\frac38y\;…….(2)substitute (2) in (1)
\begin{array}{l}\therefore y^2=4a\;\times\;\frac38y\\y^2-\frac{2ay}2=0\end{array} \begin{array}{l}\therefore y^2-\frac{3a}2y=0\\\therefore y\left(y-\frac{3a}2\right)=0\end{array} \therefore y=0\;or\;y=\frac{3a}2from equ. (2)
When y = 0, x = 0 &
When \therefore y=\frac{3a}2,\;x=\frac38\times\frac{3a}2=\frac{9a}{16}
Therefore line & parabola intersect at 0 (0,0) & A=\frac{9a}{16},\frac{3a}2
Also, differentiating (1) w.r.f. ‘y’
\begin{array}{l}2y=4a\frac{dx}{dy}\\\therefore\frac{dx}{dy}=\frac{2y}{4a}\end{array}Therefore Length of are of the parabola cut off by the line = OA
=\int_{y_1}^{y_2}\sqrt{1+\left(\frac{dx}{dy}\right)}^2dy =\int_0^{3^\frac a2}\sqrt{1+\left(\frac y{2a}\right)^2}dy =\int_0^{3^\frac a2}\sqrt{\frac{\left(2a\right)^2+y^2}{\left(2a\right)^2}}dy =\frac1{2a}\left[\frac y2\sqrt{\left(2a\right)^2+y^2}+\frac{\left(2a\right)^2}2\;\log\;\left|y+\sqrt{\left(2a\right)^2+y^2}\right|\right]_0^{3\frac a2} =\frac1{2a}\left{\left[\frac{3^{a/2}}2\sqrt{4a^2+\frac{9a^2}4}+2a^2\;\log\;\left|\frac{3a}2+\sqrt{4a^2+\frac{9^{a/2}}4-0+2a^2\;\log}\right|0+\sqrt{4a^2+0}\right]\right} =\frac1{2a}\left{\left[\frac{3a}4\sqrt{\frac{25a^2}4}+2a^2\;\log\left|\frac{3a}2+\sqrt{\frac{25a^2}4}\right|\right]-2a^2\;\log\;2a\right} =\frac1{2a}\left{\left[\frac{3a}4\times\frac{5a}2+2a^2\;\log\;\left|\frac{3a}2+\frac{5a}2\right|\right]-2a^2\;\log\;2a\right} =\frac1{2a}\times2a^2\left{\frac{15}8\times\frac12+\log\;4a-\log\;2a\right} \begin{array}{l}=a\left(\frac{15}{16}+\log\;\frac{4a}{2a}\right)\\=a\left(\log\;2+\frac{15}{16}\right)\end{array}Hence, the length of the arc of the parabola cut off by the line
3y\;=\;8x\;is\;a\left[\log\;2+\frac{15}{16}\right]Qu. 4 (B) Solve \frac{d^3y}{dx^3}-7\frac{dy}{dx}-6y=\cos\;x\;\cos h\;x
Solu. :- given \frac{d^3y}{dx^3}-7\frac{dy}{dx}-6y=\cos\;x\;\cos h\;x
It’s Auxiliary Equ is \begin{array}{l}D^3-7D-6=0\\put\;D=m\\\therefore m^3-7m-6=0\end{array}
\therefore m=3,\;-2,\;-1complimentory function (c.f.) yc
yc=c_1e^{3x}+c_2e^{-2x}+c_{3e}^{-1x}Here, X = cos x cosh x
\therefore X=\cos\;x\left(\frac{e^x+e^{-x}}2\right) \therefore X=\frac12\left(e^x\cos\;x+e^{-x}\;\cos\;x\right)Now,
Particular integral (P.I) is yp = \frac1{f\left(D\right)}x
\therefore yp=\frac1{\left(D-3\right)\left(D+2\right)\left(D+1\right)}\times\frac12\left(e^x\;\cos\;x+e^{-x}\cos\;x\right) =\frac12\left{\frac1{\left(D-3\right)\left(D+2\right)\left(D+1\right)}e^x\cos\;x+\frac1{\left(D-3\right)\left(D+2\right)\left(D+1\right)}e^{-x}\cos\;x\right} yp=\frac12\left{e^x\frac1{\left(D+1-3\right)\left(D+1+2\right)\left(D+1+1\right)}\cos\;x+e^{-x}\frac1{\left(D-1-3\right)\left(D-1+2\right)\left(D-1+1\right)}\cos\;x\right} =\frac12\left{e^x\frac1{\left(D-2\right)\left(D+3\right)\left(D+2\right)}\cos\;x+e^{-x}\frac1{\left(D-4\right)\left(D+1\right)\left(D\right)}\cos\;x\right} =\frac12\left{e^x\frac1{\left(D+3\right)\left(D^2-4\right)}\cos\;x+e^{-x}\frac1{D\left(D^2-3D-4\right)}\cos\;x\right} =\frac12\left{e^x\frac1{\left(D+3\right)\left(-1^2-4\right)}\cos\;x+e^{-x}\frac1{D.D^2-3D^2-4D}\cos\;x\right} =\frac12\left{e^x\frac1{\left(D+3\right)\left(-5\right)}\cos\;x+e^{-x}\frac1{D(-1)^2-3\times(-1)^2-4D}\cos\;x\right} =\frac12\left{e^x\frac1{-5\left(D+3\right)}\cos\;x+e^{-x}\frac1{3-5D}\cos\;x\right} =\frac12\left{e^x\frac{D-3}{-5\left(D+3\right)}\times\frac{\left(D-3\right)}{(D-3)}\cos\;x+e^{-x}\frac1{3-5D}\times\frac{\left(3+5D\right)}{\left(3+5D\right)}\cos\;x\right} yp=\frac12\left{e^x\frac{(D-3)}{-5\left(D^2-9\right)}\cos\;x+e^{-x}\frac{\left(3+5D\right)}{\left(9-25D^2\right)}\cos\;x\right} =\frac12\left{e^x\frac{D-3}{-5\left(-1^2-9\right)}\cos\;x+e^{-x}\frac{\left(3+5D\right)}{\left(9-25\times(-1)^2\right)}\cos\;x\right} =\frac12\times\frac1{50}e^x\left(D\;\cos\;x-3\;\cos\;x\right)+\frac12\times\frac1{34}e^x\left(3\;\cos\;x+5D\;\cos\;x\right) yp=\frac1{100}e^x\left(-\sin\;x-3\;\cos\;x\right)+\frac1{68}e^{-x}\left(3\;\cos\;x-5\;\sin\;x\right) \therefore The\;complete\;sol.\;is\;y\;=\;yc + yp Y=c_1e^{3x}+c_2e^{-2x}+c_3e^{-x}-\frac1{100}e^x\left(\sin\;x+3\;\cos\;x\right)+\frac1{68}e^{-x}\left(3\;\cos\;x-5\;\sin\;x\right)(C) Using four order Runge – kutta method find u(0,4) of the initial value problem u^1=2+u^2,\;u(0)=1\;take\;h=0.2\;.
Sol. :- given \begin{array}{l}u^1=2+u^2\\let,\;\;f(t,u)=\frac{du}{dt}\\=-2\;tu^2\end{array}
Here, h = 0.2, to = 0, uo = 1
Therefore By Runge kutta method of fourth order.
\begin{array}{l}k_1=hf\;(t0,\;u0)\\=0.2\;f\;(0,1)\\=0.2\;\times2\;t_0u_02\end{array} \begin{array}{l}=0.2\times2\left(0\right)\;\left(1\right)\;_2\\k_1=0\end{array} K_2=hf\left(t_0+\frac h2,\;u_0+\frac{k_1}2\right) =0.2\;f\;\left(0+\frac{0.2}2,\;1+\frac02\right) \begin{array}{l}=0.2\;f\;\left(0.1,\;1\right)\\=0.2\;\times-2\;\left(0.1\right)\;\left(1\right)^2\\K_2=0.04\end{array}Therefore The correct value of u when t = 0.2 is 0.9615
Here, h=0.2,\;t_1=0.2,\;u_1=0.9615
\begin{array}{l}K_1=hf\;\left(t_0,\;u_0\right)\\=0.2\times f\left(0.2,\;0.9615\right)\\=0.2\times-2\left(0.2\right)\;\left(0.9615\right)^2\\K_1=-0.0740\end{array} K_2=hf\left(t_0+\frac u2,\;u_0+\frac{k_1}2\right) =0.2\times f\left(0.2+\frac{0.2}2,\;0.9615-\frac{0.0740}2\right) \begin{array}{l}=0.2\times f\left(0.3,\;0.9245\right)\\=0.2\times-2\left(0.3\right)\;\left(0.9245\right)^2\\K_2=-0.1026\end{array} K_3=hf\;\left(t_0+\frac h2,\;u_0+\frac{k_2}2\right) =0.2\times f\left(0.2+\frac{0.2}2,\;0.9615-\frac{0.1026}2\right) \begin{array}{l}=0.2\times f\left(0.3,\;0.9102\right)\\=0.2\;\times-2\;\left(0.3\right)\;\left(0.9102\right)^2\\K_3=-0.0994\end{array} \begin{array}{l}K_4=hf\;\left(t_0+u,\;u_0+k_3\right)\\=0.2\times f\left(0.2+0.2,\;0.9615-0.0934\right)\\=0.2\times f\left(0.4,\;0.8621\right)\\=0.2\times-2\left(0.4\right)\;\left(0.8621\right)^2\end{array} K_4=\;-0.1189 \begin{array}{l}\therefore K=\frac16\left(k_1+2k_2+2k_3+k_4\right)\\=\frac16\left(-0.074-2\times0.1026-2\times0.0994-0.1189\right)\\K=-0.0995\end{array} \begin{array}{l}u_2=u_1+k\\=0.9615-0.0995\\u_2=0.8621\end{array}Therefore The value of u for t = 0.4 is 0.8621
i.e u(0.4) = 0.8621