Question Number 87532 by niroj last updated on 04/Apr/20
$$\:\left(\mathrm{1}\right).\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{general}}\:\boldsymbol{\mathrm{solution}}: \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{px}}\:+\boldsymbol{\mathrm{p}}^{\boldsymbol{\mathrm{n}}} \\ $$$$\:\left(\mathrm{2}\right).\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:+\:\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\:\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}\right). \\ $$$$\:\: \\ $$
Commented by mathmax by abdo last updated on 07/Apr/20
![2) (x+1)^2 y^(′′) +(x+1)y^′ =(2x+3)(2x+4) ⇒ (x+1)y^(′′) +y^′ =((4x^2 +8x+6x +12)/(x+1)) =((4x^2 +14x +12)/(x+1)) let y^′ =z (e)⇒(x+1)z^′ +z =((4x^2 +14x +12)/(x+1)) (he)→(x+1)z^′ +z =0 ⇒(x+1)z^′ =−z ⇒(z^′ /z) =−(1/(x+1)) ⇒ ln∣z∣ =−ln∣x+1∣ +c ⇒z =(k/(∣x+1∣)) let determine the solution on ]−1,+∞[ ⇒z(x) =(k/(x+1)) mvc method →z^′ =−(1/((x+1)^2 ))k +(1/(x+1))k^′ (e) ⇒−(1/(x+1))k +k^′ +(k/(x+1)) =((4x^2 +14x +12)/(x+1)) ⇒ k(x) =∫ ((4x^2 +14x +12)/(x+1))dx +c =∫ ((4(x^2 −1)+14x +16)/(x+1))dx +c =∫4(x−1) +∫((14x +16)/(x+1))dx +c =2(x−1)^2 + ∫ ((14(x+1)−14+16)/(x+1))dx +c =2(x−1)^2 +14x +2ln(x+1) +c ⇒ z(x) =((k(x))/(x+1)) =(1/(x+1))( 2x^2 −4x +2 +14x +2ln(x+1)+c) =(1/(x+1))(2x^2 +10x +2ln(x+1)+c+2) we have y^′ =z ⇒ y(x) =∫ ((2x^2 +10x +2ln(x+1)+λ)/(x+1))dx (λ=c+2)](Q87895.png)
$$\left.\mathrm{2}\right)\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} {y}^{''} \:+\left({x}+\mathrm{1}\right){y}^{'} \:=\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{4}\right)\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){y}^{''} \:+{y}^{'} \:=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{6}{x}\:+\mathrm{12}}{{x}+\mathrm{1}}\:=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{14}{x}\:+\mathrm{12}}{{x}+\mathrm{1}} \\ $$$${let}\:{y}^{'} \:={z}\:\:\:\:\left({e}\right)\Rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+{z}\:=\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{14}{x}\:+\mathrm{12}}{{x}+\mathrm{1}} \\ $$$$\left({he}\right)\rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+{z}\:=\mathrm{0}\:\Rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:=−{z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=−{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\:\:\Rightarrow{z}\:=\frac{{k}}{\mid{x}+\mathrm{1}\mid}\:{let}\:{determine}\:{the}\:{solution}\:{on} \\ $$$$\left.\right]−\mathrm{1},+\infty\left[\:\Rightarrow{z}\left({x}\right)\:=\frac{{k}}{{x}+\mathrm{1}}\:\:{mvc}\:{method}\:\rightarrow{z}^{'} \:=−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{k}\:+\frac{\mathrm{1}}{{x}+\mathrm{1}}{k}^{'} \right. \\ $$$$\left({e}\right)\:\Rightarrow−\frac{\mathrm{1}}{{x}+\mathrm{1}}{k}\:+{k}^{'} +\frac{{k}}{{x}+\mathrm{1}}\:=\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{14}{x}\:+\mathrm{12}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${k}\left({x}\right)\:=\int\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{14}{x}\:+\mathrm{12}}{{x}+\mathrm{1}}{dx}\:+{c} \\ $$$$=\int\:\frac{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{14}{x}\:+\mathrm{16}}{{x}+\mathrm{1}}{dx}\:+{c}\:=\int\mathrm{4}\left({x}−\mathrm{1}\right)\:+\int\frac{\mathrm{14}{x}\:+\mathrm{16}}{{x}+\mathrm{1}}{dx}\:+{c} \\ $$$$=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:+\:\int\:\:\frac{\mathrm{14}\left({x}+\mathrm{1}\right)−\mathrm{14}+\mathrm{16}}{{x}+\mathrm{1}}{dx}\:+{c} \\ $$$$=\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{14}{x}\:+\mathrm{2}{ln}\left({x}+\mathrm{1}\right)\:+{c}\:\Rightarrow \\ $$$${z}\left({x}\right)\:=\frac{{k}\left({x}\right)}{{x}+\mathrm{1}}\:=\frac{\mathrm{1}}{{x}+\mathrm{1}}\left(\:\:\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{2}\:+\mathrm{14}{x}\:+\mathrm{2}{ln}\left({x}+\mathrm{1}\right)+{c}\right) \\ $$$$=\frac{\mathrm{1}}{{x}+\mathrm{1}}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{10}{x}\:+\mathrm{2}{ln}\left({x}+\mathrm{1}\right)+{c}+\mathrm{2}\right) \\ $$$${we}\:{have}\:{y}^{'} ={z}\:\Rightarrow\:{y}\left({x}\right)\:=\int\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{10}{x}\:+\mathrm{2}{ln}\left({x}+\mathrm{1}\right)+\lambda}{{x}+\mathrm{1}}{dx} \\ $$$$\left(\lambda={c}+\mathrm{2}\right) \\ $$
Commented by niroj last updated on 03/May/20
thanks for effort��