Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 173507 by JordanRoddy last updated on 12/Jul/22

        solve   y−(x+1) y′ + ((2x(2x−1))/((x+1)^2 )) = 0  and      x(x^2 +x+1) y′ +((x/2) +1) y = (x^2 +x+1)^(3/2)

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${solve}\:\:\:{y}−\left({x}+\mathrm{1}\right)\:{y}'\:+\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\:\:\:{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{y}'\:+\left(\frac{{x}}{\mathrm{2}}\:+\mathrm{1}\right)\:{y}\:=\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by JordanRoddy last updated on 12/Jul/22

        solve   y−(x+1) y′ + ((2x(2x−1))/((x+1)^2 )) = 0  and      x(x^2 +x+1) y′ +((x/2) +1) y = (x^2 +x+1)^(3/2)

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${solve}\:\:\:{y}−\left({x}+\mathrm{1}\right)\:{y}'\:+\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\:\:\:{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:{y}'\:+\left(\frac{{x}}{\mathrm{2}}\:+\mathrm{1}\right)\:{y}\:=\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mokys last updated on 12/Jul/22

1) (dy/dx) − (1/(x+1))y = ((2x(2x−1))/((x+1)^3 ))    P(x) = − (1/(x+1))   , Q(x) = ((2x(2x−1))/((x+1)^3 ))     I.f = e^(∫P(x)dx)  = e^(∫ −(1/(x+1))dx) = e^(ln(1/(x+1))) = (1/(x+1))    (I.f)y = ∫ (I.f)Q(x)dx    (y/(x+1)) =2 ∫ ((x(2x−1))/((x+1)^4 )) dx    let: m = x+1 → x = m−1 →dx=dm    (y/(x+1)) = 2 ∫ (((m−1)(2m−3))/m^4 ) dm    (y/(x+1)) = 2 ∫ ((2m^2 −5m +3)/m^4 ) dm    (y/(x+1)) = 2 ∫ ((2/m^2 ) − (5/m^3 ) + (3/m^4 ))dm    (y/(x+1)) = −(4/m) + (5/m^2 ) − (2/m^3 ) + K    Now: m = x+1    ∴ y = (5/(x + 1)) − (2/(( x + 1 )^2 )) − 4 + K(x+1)    ★Mohammad Aldolaimy★

$$\left.\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:−\:\frac{\mathrm{1}}{{x}+\mathrm{1}}{y}\:=\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$${P}\left({x}\right)\:=\:−\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\:\:,\:{Q}\left({x}\right)\:=\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\:{I}.{f}\:=\:{e}^{\int{P}\left({x}\right){dx}} \:=\:{e}^{\int\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}{dx}} =\:{e}^{{ln}\frac{\mathrm{1}}{{x}+\mathrm{1}}} =\:\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$ \\ $$$$\left({I}.{f}\right){y}\:=\:\int\:\left({I}.{f}\right){Q}\left({x}\right){dx} \\ $$$$ \\ $$$$\frac{{y}}{{x}+\mathrm{1}}\:=\mathrm{2}\:\int\:\frac{{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:{dx} \\ $$$$ \\ $$$${let}:\:{m}\:=\:{x}+\mathrm{1}\:\rightarrow\:{x}\:=\:{m}−\mathrm{1}\:\rightarrow{dx}={dm} \\ $$$$ \\ $$$$\frac{{y}}{{x}+\mathrm{1}}\:=\:\mathrm{2}\:\int\:\frac{\left({m}−\mathrm{1}\right)\left(\mathrm{2}{m}−\mathrm{3}\right)}{{m}^{\mathrm{4}} }\:{dm} \\ $$$$ \\ $$$$\frac{{y}}{{x}+\mathrm{1}}\:=\:\mathrm{2}\:\int\:\frac{\mathrm{2}{m}^{\mathrm{2}} −\mathrm{5}{m}\:+\mathrm{3}}{{m}^{\mathrm{4}} }\:{dm} \\ $$$$ \\ $$$$\frac{{y}}{{x}+\mathrm{1}}\:=\:\mathrm{2}\:\int\:\left(\frac{\mathrm{2}}{{m}^{\mathrm{2}} }\:−\:\frac{\mathrm{5}}{{m}^{\mathrm{3}} }\:+\:\frac{\mathrm{3}}{{m}^{\mathrm{4}} }\right){dm} \\ $$$$ \\ $$$$\frac{{y}}{{x}+\mathrm{1}}\:=\:−\frac{\mathrm{4}}{{m}}\:+\:\frac{\mathrm{5}}{{m}^{\mathrm{2}} }\:−\:\frac{\mathrm{2}}{{m}^{\mathrm{3}} }\:+\:{K} \\ $$$$ \\ $$$${Now}:\:{m}\:=\:{x}+\mathrm{1} \\ $$$$ \\ $$$$\therefore\:{y}\:=\:\frac{\mathrm{5}}{{x}\:+\:\mathrm{1}}\:−\:\frac{\mathrm{2}}{\left(\:{x}\:+\:\mathrm{1}\:\right)^{\mathrm{2}} }\:−\:\mathrm{4}\:+\:{K}\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\bigstar{Mohammad}\:{Aldolaimy}\bigstar \\ $$

Commented by Tawa11 last updated on 13/Jul/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com