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Question Number 108498 by mohammad17 last updated on 17/Aug/20

Answered by Aziztisffola last updated on 17/Aug/20

1) y=(x+2)^x = e^(xln(x+2)) (dy/dx)=(ln(x+2)+(x/(x+2)))(x+2)^x (dy/dx)(−1)=−1

$$\left.\mathrm{1}\right)\:\mathrm{y}=\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{x}} =\:\mathrm{e}^{\mathrm{xln}\left(\mathrm{x}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{ln}\left(\mathrm{x}+\mathrm{2}\right)+\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\right)\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{x}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$

Commented by mohammad17 last updated on 17/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Aziztisffola last updated on 17/Aug/20

3) xy′+y=e^(x ) ⇒y′+(1/x)y=(e^x /x) I.F=e^(∫(dx/x)) =e^(ln∣x∣) =x yx=∫(e^x /x)xdx=e^x +C y=(e^x /x)+(C/x) 4) ch(5x)+sh(5x)=e^(5x) ∫(dx/(ch(5x)+sh(5x))) =∫ (dx/e^(5x) ) = ∫e^(−5x) dx =−(1/5) e^(−5x) +C 5) V_1 ^(→) (1;1;1) and V_2 ^(→) (1;1;−2) V_1 ^(→) .V_2 ^(→) =1+1−2 = 0 ⇒V_1 ^(→) ⊥V_2 ^(→)

$$\left.\mathrm{3}\right)\:\mathrm{xy}'+\mathrm{y}=\mathrm{e}^{\mathrm{x}\:} \Rightarrow\mathrm{y}'+\frac{\mathrm{1}}{\mathrm{x}}\mathrm{y}=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}} \\ $$$$\mathrm{I}.\mathrm{F}=\mathrm{e}^{\int\frac{\mathrm{dx}}{\mathrm{x}}} =\mathrm{e}^{\mathrm{ln}\mid\mathrm{x}\mid} =\mathrm{x} \\ $$$$\:\mathrm{yx}=\int\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}}\mathrm{xdx}=\mathrm{e}^{\mathrm{x}} +\mathrm{C} \\ $$$$\mathrm{y}=\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}}+\frac{\mathrm{C}}{\mathrm{x}} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{ch}\left(\mathrm{5x}\right)+\mathrm{sh}\left(\mathrm{5x}\right)=\mathrm{e}^{\mathrm{5x}} \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{ch}\left(\mathrm{5x}\right)+\mathrm{sh}\left(\mathrm{5x}\right)}\:=\int\:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{5x}} }\:=\:\int\mathrm{e}^{−\mathrm{5x}} \mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{e}^{−\mathrm{5x}} +\mathrm{C} \\ $$$$\left.\mathrm{5}\right)\overset{\rightarrow} {\:\mathrm{V}_{\mathrm{1}} }\left(\mathrm{1};\mathrm{1};\mathrm{1}\right)\:\mathrm{and}\overset{\rightarrow} {\:\mathrm{V}_{\mathrm{2}} }\left(\mathrm{1};\mathrm{1};−\mathrm{2}\right) \\ $$$$\overset{\rightarrow} {\:\mathrm{V}_{\mathrm{1}} }.\overset{\rightarrow} {\mathrm{V}_{\mathrm{2}} }=\mathrm{1}+\mathrm{1}−\mathrm{2}\:=\:\mathrm{0}\:\Rightarrow\overset{\rightarrow} {\mathrm{V}_{\mathrm{1}} }\bot\overset{\rightarrow} {\mathrm{V}_{\mathrm{2}} } \\ $$

Commented by mohammad17 last updated on 17/Aug/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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