1. If $ \displaystyle y = \ln (\sin^3 2x)$, then prove that If $ \displaystyle 3(\frac{{d}^{2}y}{d{x}^{2}}) + (\frac{dy}{dx})^2+36=0$.
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$ \displaystyle \begin{array}{l}\ \ \ y=\ln ({{\sin }^{3}}2x)\ \ \\\ \\\ \ \ \ \ =\ln {{(\sin 2x)}^{3}}\\\\\ \ \ \ \ =3\ln (\sin 2x)\end{array}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=\frac{3}{{\sin 2x}}\cdot \frac{d}{{dx}}(\sin 2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{3}{{\sin 2x}}\cdot \cos 2x\cdot \frac{d}{{dx}}(2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{6\cos 2x}}{{\sin 2x}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =6\cot 2x$
$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(-{{\operatorname{cosec}}^{2}}2x)\frac{d}{{dx}}(2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =-12{{\operatorname{cosec}}^{2}}2x$
$ \displaystyle \therefore 3(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}})+{{(\frac{{dy}}{{dx}})}^{2}}+36$
$ \displaystyle \begin{array}{l}=3(-12{{\operatorname{cosec}}^{2}}2x)+{{(6\cot 2x)}^{2}}+36\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\cot }^{2}}2x+1)\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\operatorname{cosec}}^{2}}2x)\ \ \ \ \left[ {1+{{{\cot }}^{2}}2x={{{\operatorname{cosec}}}^{2}}2x} \right]\\\\=0\end{array}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=\frac{3}{{\sin 2x}}\cdot \frac{d}{{dx}}(\sin 2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{3}{{\sin 2x}}\cdot \cos 2x\cdot \frac{d}{{dx}}(2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{6\cos 2x}}{{\sin 2x}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ =6\cot 2x$
$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(-{{\operatorname{cosec}}^{2}}2x)\frac{d}{{dx}}(2x)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =-12{{\operatorname{cosec}}^{2}}2x$
$ \displaystyle \therefore 3(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}})+{{(\frac{{dy}}{{dx}})}^{2}}+36$
$ \displaystyle \begin{array}{l}=3(-12{{\operatorname{cosec}}^{2}}2x)+{{(6\cot 2x)}^{2}}+36\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\cot }^{2}}2x+1)\\\\=36(-{{\operatorname{cosec}}^{2}}2x+{{\operatorname{cosec}}^{2}}2x)\ \ \ \ \left[ {1+{{{\cot }}^{2}}2x={{{\operatorname{cosec}}}^{2}}2x} \right]\\\\=0\end{array}$
2. If $ \displaystyle y = (3 + 4x) e^{-2x}$ , show that $ \displaystyle \frac{{d}^{2}y}{d{x}^{2}} + 4(\frac{dy}{dx})+4y=0$.
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$ \displaystyle \ \ \ y=(3+4x){{e}^{{-2x}}}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=(3+4x)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(3+4x)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ =(3+4x){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =-2(3+4x){{e}^{{-2x}}}+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-6-8x+4){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\end{array}$
$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=(-8x-2)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(-8x-2)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =-2(-8x-2){{e}^{{-2x}}}-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x+4-8){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x-4){{e}^{{-2x}}}\end{array}$
$ \displaystyle \therefore \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4\left( {\frac{{dy}}{{dx}}} \right)+4y$
$ \displaystyle \begin{array}{l}=(16x-4){{e}^{{-2x}}}+4(-8x-2){{e}^{{-2x}}}+4(3+4x){{e}^{{-2x}}}\\\\=(16x-4-32x-8+12+16x){{e}^{{-2x}}}\\\\=0\end{array}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=(3+4x)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(3+4x)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ =(3+4x){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =-2(3+4x){{e}^{{-2x}}}+4{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-6-8x+4){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\end{array}$
$ \displaystyle \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=(-8x-2)\cdot \frac{d}{{dx}}({{e}^{{-2x}}})+{{e}^{{-2x}}}\frac{d}{{dx}}(-8x-2)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ =(-8x-2){{e}^{{-2x}}}\frac{d}{{dx}}(-2x)-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =-2(-8x-2){{e}^{{-2x}}}-8{{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x+4-8){{e}^{{-2x}}}\\\\\ \ \ \ \ \ \ \ \ \ =(16x-4){{e}^{{-2x}}}\end{array}$
$ \displaystyle \therefore \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}+4\left( {\frac{{dy}}{{dx}}} \right)+4y$
$ \displaystyle \begin{array}{l}=(16x-4){{e}^{{-2x}}}+4(-8x-2){{e}^{{-2x}}}+4(3+4x){{e}^{{-2x}}}\\\\=(16x-4-32x-8+12+16x){{e}^{{-2x}}}\\\\=0\end{array}$
3. If $ \displaystyle y = a e ^{\sin x}$, where $ \displaystyle a$ is a constant, prove that $ \displaystyle \frac{{d}^{2}y}{d{x}^{2}} = (\cos x-\tan x)\frac{dy}{dx}$.
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$ \displaystyle \ \ \ y=a{{e}^{{\sin x}}}$
$ \displaystyle \therefore \ \frac{{dy}}{{dx}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\sin x)$
$ \displaystyle \ \ \ \ \ \ \ \ =a{{e}^{{\sin x}}}\cdot \cos x$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\cos x)+a\cos x\frac{d}{{dx}}({{e}^{{\sin x}}})$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}\cos x\frac{d}{{dx}}(\sin x)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}{{\cos }^{2}}x$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}({{\cos }^{2}}x-\sin x)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\frac{{{{{\cos }}^{2}}x}}{{\cos x}}-\frac{{\sin x}}{{\cos x}}} \right)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\cos x-\tan x} \right)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\cos x-\tan x} \right)\frac{{dy}}{{dx}}$
$ \displaystyle \therefore \ \frac{{dy}}{{dx}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\sin x)$
$ \displaystyle \ \ \ \ \ \ \ \ =a{{e}^{{\sin x}}}\cdot \cos x$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\frac{d}{{dx}}(\cos x)+a\cos x\frac{d}{{dx}}({{e}^{{\sin x}}})$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}\cos x\frac{d}{{dx}}(\sin x)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}(-\sin x)+a{{e}^{{\sin x}}}{{\cos }^{2}}x$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}({{\cos }^{2}}x-\sin x)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\frac{{{{{\cos }}^{2}}x}}{{\cos x}}-\frac{{\sin x}}{{\cos x}}} \right)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=a{{e}^{{\sin x}}}\cdot \cos x\left( {\cos x-\tan x} \right)$
$ \displaystyle \therefore \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\cos x-\tan x} \right)\frac{{dy}}{{dx}}$
4. If $ \displaystyle y=e^x(\sin 2x+ \cos 2x)$, prove that $ \displaystyle y''-2y'+5y=0$.
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$ \displaystyle \ \ \ \ y={{e}^{x}}(\sin 2x+\cos 2x)$
$ \displaystyle \therefore \ \ {y}'={{e}^{x}}\frac{d}{{dx}}(\sin 2x+\cos 2x)+(\sin 2x+\cos 2x)\frac{d}{{dx}}({{e}^{x}})$
$ \displaystyle \begin{array}{l}\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x)+{{e}^{x}}(\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x+\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}\frac{d}{{dx}}(3\cos 2x-\sin 2x)+(3\cos 2x-\sin 2x)\frac{d}{{dx}}({{e}^{x}})\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x)+{{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x+3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-7\sin 2x+\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(1)\\\\\therefore \ -2{y}'={{e}^{x}}(2\sin 2x-6\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ ------(2)\\\\\therefore \ \ 5y={{e}^{x}}(5\sin 2x+5\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(3)\\\\\text{By Equation (1) + Equation (2) + Equation (3)},\\\\\ \ \ \ {y}''-2{y}'+5y=0\end{array}$
$ \displaystyle \therefore \ \ {y}'={{e}^{x}}\frac{d}{{dx}}(\sin 2x+\cos 2x)+(\sin 2x+\cos 2x)\frac{d}{{dx}}({{e}^{x}})$
$ \displaystyle \begin{array}{l}\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x)+{{e}^{x}}(\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(2\cos 2x-2\sin 2x+\sin 2x+\cos 2x)\\\\\therefore \ \ {y}'={{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}\frac{d}{{dx}}(3\cos 2x-\sin 2x)+(3\cos 2x-\sin 2x)\frac{d}{{dx}}({{e}^{x}})\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x)+{{e}^{x}}(3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-6\sin 2x-2\cos 2x+3\cos 2x-\sin 2x)\\\\\therefore \ \ {y}''={{e}^{x}}(-7\sin 2x+\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(1)\\\\\therefore \ -2{y}'={{e}^{x}}(2\sin 2x-6\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ ------(2)\\\\\therefore \ \ 5y={{e}^{x}}(5\sin 2x+5\cos 2x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ------(3)\\\\\text{By Equation (1) + Equation (2) + Equation (3)},\\\\\ \ \ \ {y}''-2{y}'+5y=0\end{array}$
5. Find the value of $ \displaystyle x$ between $ \displaystyle 0$ and $ \displaystyle \frac{\pi }{2}$ for which the curve $ \displaystyle y = e^x \cos x$ has a stationary point. Determine whether it is a maximum or a minimum point.
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$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=-{{e}^{x}}\sin x+{{e}^{x}}\cos x$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ ={{e}^{x}}(\cos x-\sin x)\\\\\ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ {{e}^{x}}(\cos x-\sin x)=0\\\\\ \ \ \ \text{Since}\ {{e}^{x}}\ne 0\ \text{for every }x\in R,\ \\\\\ \ \ \ \cos x-\sin x=0.\\\\\therefore \ \ \cos x=\sin x\\\\\therefore \ \ \tan x=1\end{array}$
$ \displaystyle \therefore \ \ x=\frac{\pi }{4}$ [Calculus တြင္ ေထာင့္မ်ားကို တိုင္းတာရာတြင္ linear scale ျဖစ္ေသာ radian စနစ္ကိုသာ သံုးရမည္။ degree ျဖင့္ အေျဖ မေပးရပါ]
$ \displaystyle \ \ \ \ \text{When }x=\frac{\pi }{4},y={{e}^{{\frac{\pi }{4}}}}\cos \frac{\pi }{4}=\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}$
$ \displaystyle \therefore \ \ \text{The stationary point is }\left( {\frac{\pi }{4},\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}} \right).$
$ \displaystyle \left( {\frac{\pi }{4},\frac{{\sqrt{2}}}{2}{{e}^{{\frac{\pi }{4}}}}} \right)$ သည္ $ \displaystyle \left( {0.79,1.55} \right)$ ခန္႔ရွိသည္။
$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{x}}(-\sin x-\cos x)+{{e}^{x}}(\cos x-\sin x)$
$ \displaystyle \therefore \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}={{e}^{x}}(-\sin x-\cos x+\cos x-\sin x)$
$ \displaystyle \therefore \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-2{{e}^{x}}\sin x$
$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}} \right|}_{{x=\frac{\pi }{4}}}}=-2{{e}^{{\frac{\pi }{4}}}}\sin \frac{\pi }{4}=-\sqrt{2}{{e}^{{\frac{\pi }{4}}}}<0$
$ \displaystyle \therefore \ \ \text{The stationary point is a maximum turning point}$.
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