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Question Number 55668 by mr W last updated on 01/Mar/19

$$Findallfunctionsy=f\left(x\right)suchthat$$$$y^{\prime }{y}^{″}=y^{‴}.$$

Commented by malwaan last updated on 02/Mar/19

$$\mathrm{y}={\mathrm{e}}^{\mathrm{x}}$$

Commented by mr W last updated on 02/Mar/19

$$pleasechecksir:$$$$withy=e^x$$$$y^{\prime }=e^x$$$$y^{″}=e^x$$$$y^{‴}=e^x$$$$y^{\prime }{y}^{″}=e^x{e}^x=e^{2x}\ne y^{{}^{‴}}=e^x$$$$\Rightarrow y=e^{x}isnotasolution$$

Commented by mr W last updated on 02/Mar/19

$${it}^{\prime }sobviousthaty=c_{1}x+c_{2}isasolution,$$$$since{y}^{\prime }=c_1,y^{″}=0,y^{‴}=0\Rightarrow y^{\prime }{y}^{″}=y^{‴}=0.$$$$butarethereanyothersolutions?$$

Answered by mr W last updated on 03/Mar/19

$$letg\left(x\right)=y^{\prime }=f^{\prime }\left(x\right)$$$$\Rightarrow {gg}^{\prime }=g^{″}$$$$leth\left(x\right)=g^{\prime }\left(x\right)$$$$g^{″}=\frac{dh\left(x\right)}{dx}=\frac{dh\left(x\right)}{dg}\times \frac{dg}{dx}=h\frac{dh}{dg}$$$$\Rightarrow gh=h\frac{dh}{dg}$$$$ifh\ne 0:(h=0\Rightarrow g\left(x\right)=c_1\Rightarrow y=c_{1}x+c_2)$$$$\Rightarrow gdg=dh$$$$\Rightarrow \int gdg=\int dh$$$$\Rightarrow \frac{g^2+c_1}{2}=h=\frac{dg}{dx}$$$$\Rightarrow (g^2+c_1)dx=2dg$$$$\mathit{i}\mathit{f}{\mathit{g}}^2+{\mathit{c}}_1\ne 0:$$$$\Rightarrow \frac{dg}{g^2+c_1}=\frac{dx}{2}$$$$\Rightarrow \int \frac{dg}{g^2+c_1}=\int \frac{dx}{2}$$$$if{c}_1=0:$$$$\Rightarrow \int \frac{dg}{g^2}=\int \frac{dx}{2}$$$$\Rightarrow -\frac{1}{g}=\frac{x+c_2}{2}$$$$\Rightarrow g=y^{\prime }=\frac{dy}{dx}=-\frac{2}{x+c_2}$$$$\Rightarrow dy=-\frac{2dx}{x+c_2}$$$$\Rightarrow y=-2\ln \mid x+c_2\mid +c_3$$$$if{c}_1>0,say{c}_1=c_4^2$$$$\Rightarrow \int \frac{dg}{g^2+c_4^2}=\int \frac{dx}{2}$$$$\Rightarrow \frac{1}{c_4}{\tan }^{-1}\frac{g}{c_4}=\frac{x+c_5}{2}$$$$\Rightarrow g=y^{\prime }=\frac{dy}{dx}=c_4\tan \frac{c_4(x+c_5)}{2}$$$$\Rightarrow y=\int c_4\tan \frac{c_4(x+c_5)}{2}dx$$$$\Rightarrow y=2\int \tan \frac{c_4(x+c_5)}{2}d\left(\frac{c_4(x+c_5)}{2}\right)$$$$\Rightarrow y=-2\ln \mid \cos \frac{c_4(x+c_5)}{2}\mid +c_6$$$$if{c}_1<0,say{c}_1=-c_4^2$$$$\Rightarrow \int \frac{dg}{g^2-c_4^2}=\int \frac{dx}{2}$$$$\Rightarrow \frac{1}{2c_4}\ln \mid \frac{g-c_4}{g+c_4}\mid =\frac{x+c_5}{2}$$$$\Rightarrow \ln \mid 1-\frac{2c_4}{g+c_4}\mid =c_4(x+c_5)$$$$\Rightarrow 1-\frac{2c_4}{g+c_4}=e^{c_4(x+c_5)}$$$$\Rightarrow \frac{2c_4}{g+c_4}=1-e^{c_4(x+c_5)}$$$$\Rightarrow g=y^{\prime }=\frac{dy}{dx}=c_4[\frac{2}{1-e^{c_4(x+c_5)}}-1]$$$$\Rightarrow y=\int c_4[\frac{2}{1-e^{c_4(x+c_5)}}-1]dx$$$$\Rightarrow y=c_4[\int \frac{2dx}{1-e^{c_4(x+c_5)}}-x]$$$$\Rightarrow y=c_4[2(x+c_5)-\frac{2\ln \mid 1-e^{c_4(x+c_5)}\mid }{c_4}-x]$$$$\Rightarrow y=c_{4}x-2\ln \mid 1-e^{c_4(x+c_5)}\mid +c_6$$$$\mathit{i}\mathit{f}{\mathit{g}}^2+{\mathit{c}}_1=0:$$$$if{c}_1=0:$$$$\Rightarrow g=y^{\prime }=\frac{dy}{dx}=0$$$$\Rightarrow y=c_2$$$$if{c}_1>0:g^2+c_1\ne 0$$$$if{c}_1<0:g^2+c_1=0$$$$\Rightarrow g^2=-c_1=c_2^{2}say$$$$\Rightarrow g=\pm c_2$$$$\Rightarrow g=\frac{dy}{dx}=\pm c_2$$$$\Rightarrow y=\pm c_{2}x+c_3$$$$$$$$summary:$$$$followingfunctionsarepossiblesolutions:$$$$y=f\left(x\right)=c_{1}x+c_2$$$$y=f\left(x\right)=-2\ln \mid x+c_1\mid +c_2$$$$y=f\left(x\right)=-2\ln \mid \cos (c_{1}x+c_2)\mid +c_3$$$$y=f\left(x\right)=-2\ln \mid 1-e^{c_{1}x+c_2}\mid +c_{1}x+c_3$$

Commented by MJS last updated on 02/Mar/19

$$\mathrm{great}!$$

Commented by otchereabdullai@gmail.com last updated on 02/Mar/19

$$\mathrm{Ideal}\mathrm{Prof}$$

Commented by malwaan last updated on 03/Mar/19

$$\mathrm{fantastic}\mathrm{proof}$$

Commented by rahul 19 last updated on 03/Mar/19

$$Wonderful!$$

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