3FEBIT-MATHEMATICS 2-ASS1-2Use the following information for Questions 10 and 11:2ty =Y+4.t10.The derivative of the left-hand side of the equation with respect to t is:11.The derivative of the right-hand side of the equation with respect to t is:12. The second-degree Maclaurin polynomial for f(x)= vx+9 is 3+x2If this6 108polynomial is used to approximate v8, the answer is:dy2- 3t3+2tis1+t13.of the parametric equations x =and ydx1+tUse the following information for Questions 14 and 15:

We have to evaluate e2ty=yt+4 ⇒e2ty-yt-4=0 Thus, we have given fx=e2ty-yt-4 10 Now, we have to…

Answered: Use the following information for…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Expert Solution

Step 1

We have to evaluate $e^{2 t y} = \frac{y}{t} + 4$

$\Rightarrow e^{2 t y} - \frac{y}{t} - 4 = 0$

Thus, we have given $f (x) = e^{2 t y} - \frac{y}{t} - 4$

$(10)$ Now, we have to evaluate Left hand derivative:

we know that $L H D = \lim_{h \to 0} \frac{f (c - h) - f (c)}{- h}$

Now,

$f' (t) = \lim_{h \to 0} \frac{f (t - h) - f (t)}{- h} f' (t) = \lim_{h \to 0} \frac{(e^{2 (t - h) y} - \frac{y}{(t - h)} - 4) - (e^{2 t y} - \frac{y}{t} - 4)}{- h} f' (t) = \lim_{h \to 0} \frac{e^{2 (t - h) y} - \frac{y}{(t - h)} - e^{2 t y} + \frac{y}{t}}{- h} f' (t) = \lim_{h \to 0} \frac{t (t - h) e^{2 (t - h) y} - t y - t (t - h) e^{2 t y} + y (t - h)}{- h t (t - h)} f' (t) = \lim_{h \to 0} \frac{(t^{2} - t h) e^{2 (t - h) y} - t y - (t^{2} - t h) e^{2 t y} + y t - y h}{- h t (t - h)} f' (t) = \lim_{h \to 0} \frac{(t^{2} - t h) e^{2 t y} e^{- 2 h y} - (t^{2} - t h) e^{2 t y} - y h}{- h t (t - h)} f' (t) = \lim_{h \to 0} \frac{(t^{2} - t h) e^{2 t y} (e^{- 2 h y} - 1) - y h}{- h t (t - h)} f' (t) = \lim_{h \to 0} \frac{(t^{2} - t h) e^{2 t y} (e^{- 2 h y} - 1)}{- h t (t - h)} - \frac{y h}{- h t (t - h)} f' (t) = \lim_{h \to 0} \frac{2 y (t^{2} - t h) e^{2 t y} (e^{- 2 h y} - 1)}{- 2 y h t (t - h)} - \frac{y h}{- h t (t - h)} (∵ \lim_{h \to 0} \frac{e^{- 2 h y} - 1}{- 2 h y} = 1) f' (t) = \lim_{h \to 0} 2 y e^{2 t y} + \frac{y}{t (t - h)} f' (t) = 2 y e^{2 t y} + \frac{y}{t^{2}}$