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Question Number 83639 by niroj last updated on 04/Mar/20

Find the differential equations: (i) log((dy/dx))=ax+by (ii) x cos y dy=(x e^x log x +e^x )dx

$$ \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equations}}: \\ $$$$\:\:\:\left(\mathrm{i}\right)\:\mathrm{log}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{ax}+\mathrm{by} \\ $$$$\:\:\:\left(\mathrm{ii}\right)\:\mathrm{x}\:\mathrm{cos}\:\mathrm{y}\:\mathrm{dy}=\left(\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \mathrm{log}\:\mathrm{x}\:+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 05/Mar/20

1)ln((dy/dx))=ax+by (dy/dx)=e^(ax+by) (dy/e^(by) )=e^(ax) dx ∫e^(−by) dy=∫e^(ax) dx (e^(−by) /(−b))=(e^(ax) /a)+c

$$\left.\mathrm{1}\right){ln}\left(\frac{{dy}}{{dx}}\right)={ax}+{by} \\ $$$$\frac{{dy}}{{dx}}={e}^{{ax}+{by}} \\ $$$$\frac{{dy}}{{e}^{{by}} }={e}^{{ax}} {dx} \\ $$$$\int{e}^{−{by}} {dy}=\int{e}^{{ax}} {dx} \\ $$$$\frac{{e}^{−{by}} }{−{b}}=\frac{{e}^{{ax}} }{{a}}+{c} \\ $$

Commented by niroj last updated on 05/Mar/20

its ★ great

$${its}\:\bigstar\:{great} \\ $$

Commented by TANMAY PANACEA last updated on 05/Mar/20

most welcome

$${most}\:{welcome}\: \\ $$

Answered by TANMAY PANACEA last updated on 05/Mar/20

2)xcosydy−(xe^x lnx+e^x )dx ∫cosydy−∫(e^x lnx+(e^x /x))dx=0 ∫cosydy−∫(d/dx)(e^x lnx)dx=0 siny−e^x lnx=c

$$\left.\mathrm{2}\right){xcosydy}−\left({xe}^{{x}} {lnx}+{e}^{{x}} \right){dx} \\ $$$$\int{cosydy}−\int\left({e}^{{x}} {lnx}+\frac{{e}^{{x}} }{{x}}\right){dx}=\mathrm{0} \\ $$$$\int{cosydy}−\int\frac{{d}}{{dx}}\left({e}^{{x}} {lnx}\right){dx}=\mathrm{0} \\ $$$${siny}−{e}^{{x}} {lnx}={c} \\ $$

Commented by niroj last updated on 05/Mar/20

thank you panacea.

$$\:{thank}\:{you}\:{panacea}. \\ $$

Commented by TANMAY PANACEA last updated on 05/Mar/20

most welcome

$${most}\:{welcome} \\ $$

Answered by M±th+et£s last updated on 05/Mar/20

log((dy/dx))=ax+by→log(p)=ax+by→y=(1/b)(log(p)−ax) Diff. w.r.t. x→p=(1/b)(((d log(p))/dx) −a) p=(1/b)((1/p) (dp/dx)−a)→(dx/dp)=(1/(p(a+bp)))=(1/a) (1/p)−(b/a) (1/(a+bp)) x=(1/a)ln∣p∣−(1/a)ln∣a+bp∣+(1/a)ln(c) x=(1/a)ln(((cp)/(a+bp)))→e^(ax) =((cp)/(a+bp))→p=((ae^(ax) )/(c−be^(ax) )) y= −((ln(c−be^(ax) ))/b)

$${log}\left(\frac{{dy}}{{dx}}\right)={ax}+{by}\rightarrow{log}\left({p}\right)={ax}+{by}\rightarrow{y}=\frac{\mathrm{1}}{{b}}\left({log}\left({p}\right)−{ax}\right) \\ $$$${Diff}.\:{w}.{r}.{t}.\:{x}\rightarrow{p}=\frac{\mathrm{1}}{{b}}\left(\frac{{d}\:{log}\left({p}\right)}{{dx}}\:−{a}\right) \\ $$$${p}=\frac{\mathrm{1}}{{b}}\left(\frac{\mathrm{1}}{{p}}\:\frac{{dp}}{{dx}}−{a}\right)\rightarrow\frac{{dx}}{{dp}}=\frac{\mathrm{1}}{{p}\left({a}+{bp}\right)}=\frac{\mathrm{1}}{{a}}\:\frac{\mathrm{1}}{{p}}−\frac{{b}}{{a}}\:\frac{\mathrm{1}}{{a}+{bp}} \\ $$$${x}=\frac{\mathrm{1}}{{a}}{ln}\mid{p}\mid−\frac{\mathrm{1}}{{a}}{ln}\mid{a}+{bp}\mid+\frac{\mathrm{1}}{{a}}{ln}\left({c}\right) \\ $$$${x}=\frac{\mathrm{1}}{{a}}{ln}\left(\frac{{cp}}{{a}+{bp}}\right)\rightarrow{e}^{{ax}} =\frac{{cp}}{{a}+{bp}}\rightarrow{p}=\frac{{ae}^{{ax}} }{{c}−{be}^{{ax}} } \\ $$$${y}=\:−\frac{{ln}\left({c}−{be}^{{ax}} \right)}{{b}} \\ $$$$ \\ $$$$ \\ $$

Answered by M±th+et£s last updated on 05/Mar/20

cos(y)dy=((xe^x log(x)+e^x )/x)dx →sin(y)=∫(e^x log(x)+(e^x /x))dx=∫e^x log(x)dx+e^x log(x)−∫e^x log(x) dx sin(y)=e^x log(x)+c→y=sin^(−1) (c+e^x log(x))

$${cos}\left({y}\right){dy}=\frac{{xe}^{{x}} \:{log}\left({x}\right)+{e}^{{x}} }{{x}}{dx} \\ $$$$\rightarrow{sin}\left({y}\right)=\int\left({e}^{{x}} {log}\left({x}\right)+\frac{{e}^{{x}} }{{x}}\right){dx}=\int{e}^{{x}} {log}\left({x}\right){dx}+{e}^{{x}} {log}\left({x}\right)−\int{e}^{{x}} \:{log}\left({x}\right)\:{dx} \\ $$$${sin}\left({y}\right)={e}^{{x}} {log}\left({x}\right)+{c}\rightarrow{y}={sin}^{−\mathrm{1}} \left({c}+{e}^{{x}} {log}\left({x}\right)\right) \\ $$$$ \\ $$

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