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Question Number 107729 by  M±th+et+s last updated on 12/Aug/20

$$1)solvetheD.E:$$$$\frac{dy}{dx}-ytan\left(x\right)=-y^{2}sec\left(x\right)$$$$$$$$2)findx$$$$\underset{k=1}{\prod ^{25}}(x+\frac{x}{3^k})=1$$$$$$$$3)solve:$$$$\mid Z^2\mid -4Z=0$$

Commented by mohammad17 last updated on 12/Aug/20

$$$$$$1)y^{-2}\frac{dy}{dx}-y^{-1}tanx=-secx$$$$$$$$let:y^{-1}=w\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dw}{dx}$$$$$$$$\frac{dw}{dx}+wtanx=secx\Rightarrow p\left(x\right)=tanx,Q\left(x\right)=secx$$$$$$$$(I.f)=e^{\int p\left(x\right)dx}=e^{\int tanxdx}=e^{-ln\mid sinx\mid }=\frac{1}{sinx}$$$$$$$$w=\frac{\int (I.f)Q\left(x\right)dx}{(I.f)}\Rightarrow w=\frac{\int secxcscxdx}{cscx}$$$$$$$$w=\frac{2\int csc2x\times \frac{csc2x+cot2x}{csc2x+cot2x}dx}{cscx}\Rightarrow w=\frac{2\int \frac{{csc}^{2}2x+csc2xcot2x}{csc2x+cot2x}dx}{csc2x}$$$$$$$$w=\frac{-ln\mid csc2x+cot2x\mid +c}{csc2x}\Rightarrow \frac{1}{y}=\frac{-ln\mid csc2x+cot2x\mid +c}{csc2x}$$$$$$$$y=\frac{csc2x}{c-ln\mid csc2x+cot2x\mid }$$$$$$$$by:mss.mohammad$$

Commented by Dwaipayan Shikari last updated on 12/Aug/20

$$\frac{dy}{dx}-\frac{1}{v}tanx=-\frac{1}{v^2}secxy=\frac{1}{v},\frac{dy}{dx}=-\frac{1}{v^2}\frac{dv}{dx}$$$$-\frac{1}{v^2}.\frac{dv}{dx}-\frac{1}{v}tanx=-\frac{1}{v^2}secx$$$$\frac{dv}{dx}+vtanx=secx$$$$I.F=e^{\int tanx}=secx$$$$v.secx=\int {sec}^{2}xdx$$$$v=\frac{tanx}{secx}+Ccosx$$$$\frac{1}{y}=sinx+Ccosx$$$$y=\frac{1}{sinx+Ccosx}$$

Commented by Dwaipayan Shikari last updated on 12/Aug/20

$$\mid x^2-y^2+2ixy\mid -4x-4iy=0(Z=x+iy)$$$$\sqrt{{(x^2-y^2)}^2+4x^2{y}^2}-4(x+iy)=0$$$$x^2+y^2=4x+4iy$$$$4iy=0\Rightarrow y=0$$$$x^2+y^2=4x$$$$x(x-4)=0$$$$x=4orx=0$$$$Z=4orZ=0$$$$$$

Answered by Her_Majesty last updated on 12/Aug/20

$$3)$$$$Z={re}^{i\theta };r>0$$$$r^2=4{re}^{i\theta }$$$$r=4e^{i\theta }$$$$r=4cos\theta +4isin\theta$$$$\Rightarrow$$$$4isin\theta =0\iff \theta =n\pi$$$$\Rightarrow Z=4e^{in\pi }=4becauser>0$$$$...andofcourse,Z=0istheobvioussolution$$

Commented by  M±th+et+s last updated on 12/Aug/20

$$thankyousir$$

Answered by abdomathmax last updated on 12/Aug/20

$$1){\mathrm{y}}^{{}^{\prime }}-\mathrm{ytanx}=-\frac{{\mathrm{y}}^2}{\mathrm{cosx}}\Rightarrow -\frac{{\mathrm{y}}^{{}^{\prime }}}{{\mathrm{y}}^2}+\frac{1}{\mathrm{y}}\mathrm{tanx}=\frac{1}{\mathrm{cosx}}$$$$\mathrm{let}\mathrm{z}=\frac{1}{\mathrm{y}}\Rightarrow {\mathrm{z}}^{{}^{\prime }}=-\frac{{\mathrm{y}}^{{}^{\prime }}}{{\mathrm{y}}^2}\mathrm{so}\mathrm{e}\Rightarrow {\mathrm{z}}^{{}^{\prime }}+\left(\mathrm{tanx}\right)\mathrm{z}=\frac{1}{\mathrm{cosx}}$$$$\mathrm{h}\to {\mathrm{z}}^{{}^{\prime }}=-\mathrm{z}\mathrm{tanx}\Rightarrow \frac{{\mathrm{z}}^{{}^{\prime }}}{\mathrm{z}}=-\mathrm{tanx}\Rightarrow \ln \mid \mathrm{z}\mid =-\int \frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}$$$$=\ln \mid \mathrm{cosx}\mid +\mathrm{c}\Rightarrow \mathrm{z}=\mathrm{k}\mid \mathrm{cosx}\mid \mathrm{solution}\mathrm{on}$$$$\{\mathrm{x}/\mathrm{cosx}>0\}\mathrm{mvc}\mathrm{method}{\mathrm{z}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\mathrm{cosx}-\mathrm{ksinx}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}\mathrm{cosx}-\mathrm{ksinx}+\mathrm{ksinx}=\frac{1}{\mathrm{cosx}}\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}=\frac{1}{{\cos }^2\mathrm{x}}\Rightarrow \mathrm{k}=\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}=\mathrm{tanx}+\lambda \Rightarrow$$$$\mathrm{z}=(\mathrm{tanx}+\lambda )\mathrm{cosx}=\mathrm{sinx}+\lambda \mathrm{cosx}\Rightarrow$$$$\mathrm{y}=\frac{1}{\mathrm{sinx}+\lambda \mathrm{cosx}}$$

Commented by  M±th+et+s last updated on 12/Aug/20

$$welldonesir$$

Commented by abdomathmax last updated on 12/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Commented by Ar Brandon last updated on 12/Aug/20

Est-ce la méthode de variation de constante, monsieur ?��

Commented by abdomathmax last updated on 12/Aug/20

$$\mathrm{oui}$$

Commented by Ar Brandon last updated on 12/Aug/20

OK��

Commented by mohammad17 last updated on 12/Aug/20

$$sirwhatsthemeanmvcmethodcanyougivemeinformationaboutthis$$

Commented by Ar Brandon last updated on 12/Aug/20

It means variation of parameters in English.

Commented by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{yes}$$

Commented by Ar Brandon last updated on 12/Aug/20

I would like to master the mvc and the wronskian method. But I haven't found any reliable document concerning these subjects. Can someone please help me ? ��

Commented by Ar Brandon last updated on 12/Aug/20

$$\mathrm{Sir}\mathrm{can}\mathrm{you}\mathrm{use}\mathrm{your}\mathrm{mvc}\mathrm{method}\mathrm{to}\mathrm{prove}\mathrm{that}\mathrm{the}$$$$\mathrm{answer}\mathrm{to}\mathrm{Q}107727\mathrm{is}\mathrm{x}\left(\mathrm{t}\right)=\frac{1}{\lambda {(1+{\mathrm{t}}^3)}^{\frac{1}{3}}-2};\lambda \in \mathbb{R}?$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{arrive}\mathrm{there}.$$

Answered by abdomathmax last updated on 12/Aug/20

$$3)\mathrm{let}\mathrm{Z}=\mathrm{x}+\mathrm{iy}$$$$\mathrm{e}\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4(\mathrm{x}+\mathrm{iy})=0\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{iy}=0\Rightarrow$$$$\{\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}=0\\ \mathrm{y}=0\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}(\mathrm{x}-4)=0\\ \mathrm{y}=0\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}=0\mathrm{or}\mathrm{x}=4\\ \mathrm{y}=0\end{array}$$$$\Rightarrow \mathrm{Z}=0\mathrm{or}\mathrm{Z}=4$$

Answered by bemath last updated on 12/Aug/20

$$\left(1\right)\frac{dy}{dx}-y\tan x=-y^2\sec x$$$$letv=y^{1-2}=y^{-1}$$$$\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-y^2\frac{dv}{dx}$$$$\Rightarrow -y^2\frac{dv}{dx}-y\tan x=-y^2\sec x$$$$\frac{dv}{dx}+\frac{\tan x}{y}=\sec x$$$$\frac{dv}{dx}+v.\tan x=\sec x$$$$takingintegratingfactor$$$$u\left(x\right)=e^{\int \tan xdx}=e^{\ln \left(\sec x\right)}=\sec x$$$$multiplybothsideby\sec x$$$$\Rightarrow \sec x\frac{dv}{dx}+v.\sec x.\tan x=\sec {}^{2}x$$$$\frac{d}{dx}(v.\sec x)=\sec {}^{2}x\Rightarrow \int d(v.\sec x)=\tan x+C$$$$v.\sec x=\tan x+C$$$$\frac{\sec x}{y}=\tan x+C$$$$\frac{y}{\sec x}=\frac{1}{\tan x+C}$$$$y=\frac{\sec x}{\tan x+C}=\frac{1}{\sin x+C.\cos x}$$

Commented by bobhans last updated on 12/Aug/20

$$\mathrm{i}\mathrm{like}\mathrm{Bernoulli}\mathrm{diff}\mathrm{equation}.$$

Commented by john santu last updated on 12/Aug/20

$$Bernoulli....great.....andcooll$$

Answered by Ar Brandon last updated on 12/Aug/20

$$1.\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{ytanx}=-{\mathrm{y}}^2\mathrm{secx}$$$$\Rightarrow -\frac{1}{{\mathrm{y}}^2}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{y}}\mathrm{tanx}=\mathrm{secx}$$$$\mathrm{Let}\mathrm{u}=\frac{1}{\mathrm{y}}\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=-\frac{1}{{\mathrm{y}}^2}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}$$$$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{u}\mathrm{tanx}=\mathrm{secx}$$$$\mathrm{IF};\varphi \left(\mathrm{x}\right)={\mathrm{e}}^{\int \mathrm{tanxdx}}=\mathrm{secx}$$$$\Rightarrow \mathrm{d}\left(\mathrm{u}\mathrm{secx}\right)=\left({\sec }^2\mathrm{x}\right)\mathrm{dx}$$$$\Rightarrow \mathrm{u}\mathrm{secx}=\int {\sec }^2\mathrm{xdx}=\mathrm{tanx}+\lambda$$$$\Rightarrow \frac{1}{\mathrm{y}}=\mathrm{sinx}+\lambda \mathrm{cosx}$$$$\Rightarrow \mathrm{y}=\frac{1}{\mathrm{sinx}+\lambda \mathrm{cosx}}$$

Answered by Ar Brandon last updated on 12/Aug/20

$$2.\underset{\mathrm{k}=1}{\prod ^{25}}(\mathrm{x}+\frac{\mathrm{x}}{3^{\mathrm{k}}})=1\Rightarrow {\mathrm{x}}^{25}\underset{\mathrm{k}=1}{\prod ^{25}}(1+\frac{1}{3^{\mathrm{k}}})=1$$$$\Rightarrow \mathrm{x}={\left[\underset{\mathrm{k}=1}{\prod ^{25}}(1+\frac{1}{3^{\mathrm{k}}})\right]}^{-\frac{1}{25}}=\underset{\mathrm{k}=1}{\prod ^{25}}{[1+\frac{1}{3^{\mathrm{k}}}]}^{-\frac{1}{25}}$$

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