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Question Number 67759 by ugwu Kingsley last updated on 31/Aug/19

$$usingvariationofparametersmethod$$$$$$$${(x+2)}^2{y}^{″}-(x+2)y^{\prime }=2x+4$$$$$$$$$$$$x^2{y}^{″}+2{xy}^{\prime }-2y=x^{2}lnx+3x$$

Commented by mathmax by abdo last updated on 31/Aug/19

$$1)changement{y}^{{}^{\prime }}=zgive{(x+2)}^2{z}^{\prime }-(x+2)z=2x+4$$$$\left(he\right)\to {(x+2)}^2{z}^{{}^{\prime }}-(x+2)z=0\Rightarrow {(x+2)}^2{z}^{{}^{\prime }}=(x+2)z\Rightarrow$$$$\frac{z^{{}^{\prime }}}{z}=\frac{x+2}{{(x+2)}^2}\Rightarrow \frac{z^{{}^{\prime }}}{z}=\frac{1}{x+2}\Rightarrow ln\mid z\mid =ln\mid x+2\mid +c\Rightarrow z=K\mid x+2\mid letfind$$$$thesolutionon]-2,+\mathrm{\infty }[letusemvcmethod\Rightarrow z^{{}^{\prime }}=K^{{}^{\prime }}(x+2)+K$$$$\left(e\right)\Rightarrow {(x+2)}^2{K}^{{}^{\prime }}(x+2)+K{(x+2)}^2-(x+2)K(x+2)=2x+4\Rightarrow$$$${(x+2)}^3{K}^{{}^{\prime }}=2(x+2)\Rightarrow K^{{}^{\prime }}=\frac{2}{{(x+2)}^2}\Rightarrow K\left(x\right)=-\frac{2}{x+2}+c\Rightarrow$$$$z\left(x\right)=(-\frac{2}{x+2}+c)(x+2)=-2+c$$$$y^{{}^{\prime }}=z\Rightarrow y\left(x\right)=\int z\left(x\right)dx+\lambda =\int (-2+c)dx+\lambda$$$$=(-2+c)x+\lambda$$

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