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use-bernoulli-methods-sec-2-y-dy-dx-xtany-x-3-




Question Number 34724 by mondodotto@gmail.com last updated on 10/May/18
use bernoulli methods  sec^2 y(dy/dx)+xtany=x^3
$$\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{bernoulli}}\:\boldsymbol{\mathrm{methods}} \\ $$$$\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{{y}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\boldsymbol{{x}\mathrm{tan}{y}}=\boldsymbol{{x}}^{\mathrm{3}} \\ $$
Commented by mondodotto@gmail.com last updated on 10/May/18
please someone solve this
$$\mathrm{please}\:\mathrm{someone}\:\mathrm{solve}\:\mathrm{this}\: \\ $$
Answered by candre last updated on 10/May/18
v=tan y  (dv/dx)=sec^2 y(dy/dx)  (dv/dx)+xv=x^3   μ(x)=e^(∫xdx) =e^(x^2 /2)   e^(x^2 /2) v′+xe^(x^2 /2) v=e^(x^2 /2) x^3   ∫(e^(x^2 /2) v)′dx=∫x^3 e^(x^2 /2) dx  e^(x^2 /2) v=e^(x^2 /2) (x^2 −2)+C  v=x^2 −2+Ce^(−x^2 /2)   tan y=x^2 −2+Ce^(−x^2 /2)
$${v}=\mathrm{tan}\:{y} \\ $$$$\frac{{dv}}{{dx}}=\mathrm{sec}^{\mathrm{2}} {y}\frac{{dy}}{{dx}} \\ $$$$\frac{{dv}}{{dx}}+{xv}={x}^{\mathrm{3}} \\ $$$$\mu\left({x}\right)={e}^{\int{xdx}} ={e}^{{x}^{\mathrm{2}} /\mathrm{2}} \\ $$$${e}^{{x}^{\mathrm{2}} /\mathrm{2}} {v}'+{xe}^{{x}^{\mathrm{2}} /\mathrm{2}} {v}={e}^{{x}^{\mathrm{2}} /\mathrm{2}} {x}^{\mathrm{3}} \\ $$$$\int\left({e}^{{x}^{\mathrm{2}} /\mathrm{2}} {v}\right)'{dx}=\int{x}^{\mathrm{3}} {e}^{{x}^{\mathrm{2}} /\mathrm{2}} {dx} \\ $$$${e}^{{x}^{\mathrm{2}} /\mathrm{2}} {v}={e}^{{x}^{\mathrm{2}} /\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{2}\right)+{C} \\ $$$${v}={x}^{\mathrm{2}} −\mathrm{2}+{Ce}^{−{x}^{\mathrm{2}} /\mathrm{2}} \\ $$$$\mathrm{tan}\:{y}={x}^{\mathrm{2}} −\mathrm{2}+{Ce}^{−{x}^{\mathrm{2}} /\mathrm{2}} \\ $$
Commented by candre last updated on 10/May/18
sec^2 y(dy/dx)=2x−Cxe^(−x^2 /2)   xtan y=x^3 −2x+Cxe^(−x^2 /2)   sec^2 y(dy/dx)+xtan y=x^3
$$\mathrm{sec}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}=\mathrm{2}{x}−{Cxe}^{−{x}^{\mathrm{2}} /\mathrm{2}} \\ $$$${x}\mathrm{tan}\:{y}={x}^{\mathrm{3}} −\mathrm{2}{x}+{Cxe}^{−{x}^{\mathrm{2}} /\mathrm{2}} \\ $$$$\mathrm{sec}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}+{x}\mathrm{tan}\:{y}={x}^{\mathrm{3}} \\ $$

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