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Question Number 81824 by M±th+et£s last updated on 15/Feb/20

Answered by mind is power last updated on 17/Feb/20

$$letln\left(x\right)=t$$$$\Rightarrow y\left(x\right)=y\left(e^t\right)=z\left(t\right)$$$$\frac{dy}{dx}=z^{\prime }\left(t\right).\frac{1}{x},\frac{d^{2}y}{{dx}^2}=-\frac{z^{\prime }\left(t\right)}{x^2}+\frac{z^{″}\left(t\right)}{x^2}$$$$\frac{d^{3}y}{{dx}^3}=-\frac{2}{x^3}(-z^{\prime }+z^{″})+\frac{1}{x^3}(-z^{″}+z^{‴})$$$$=\frac{z^{‴}}{x^3}-3\frac{z^{″}}{x^3}+\frac{2z^{\prime }}{x^3}$$$$x^3{y}^{‴}+3x^2{y}^{″}-3{xy}^{\prime }=4sh(ln\left(x\right)ch\left(ln\left(x\right)\right)$$$$\iff z^{‴}-3z^{″}+2z^{\prime }+3(-z^{\prime }+z^{″})-3z^{\prime }=2sh\left(2t\right)$$$$\Rightarrow z^{‴}-4z^{\prime }=2sh\left(2t\right)$$$$z^{\prime }=u$$$$\iff u^{″}-4u=2sh\left(2t\right)=e^{2t}+e^{-2t}$$$$u\left(t\right)={ae}^{2t}+{be}^{-2t}...homgenius$$$$u_p=\frac{1}{4}{te}^{2t}-\frac{1}{4}{te}^{-2t}particularsolutio7$$$$u={ae}^{2t}+{be}^{-2t}+\frac{t}{2}sh\left(2t\right)$$$$z=\int udt={ce}^{2t}+{de}^{-2t}+\frac{t}{4}ch\left(2t\right)-\frac{sh\left(2t\right)}{8}+c$$$$z\left(ln\left(x\right)\right)=y\left(x\right)$$$$y\left(x\right)={cx}^2+{dx}^{-2}+\frac{ln\left(x\right)}{4}\left(\frac{x^2-x^{-2}}{2}\right)-\frac{x^2-x^{-2}}{8}+c$$$$y\left(x\right)={ex}^2+{fx}^{-2}+\frac{ln\left(x\right)}{8}(x^2-x^{-2})+r$$$$\overline{ef},rareconstant$$

Commented by M±th+et£s last updated on 17/Feb/20

$$godblessyou$$

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