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

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