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Question Number 162959 by mkam last updated on 02/Jan/22

$$solvethedifferentialequationy=x+p^3$$

Commented by mr W last updated on 02/Jan/22

$$thisisnotadifferentialequation.$$

Commented by Kamel last updated on 02/Jan/22

$$MrWp=\frac{dy}{dx}.$$

Commented by mr W last updated on 02/Jan/22

$$whatisthen{p}^3?$$$$p^3={\left(\frac{dy}{dx}\right)}^3?or{p}^3=y^{‴}=\frac{d^{3}y}{{dx}^3}?$$

Commented by Kamel last updated on 02/Jan/22

$$\left(E\right)\iff {\left(\frac{dy}{dx}\right)}^3+x=y$$

Commented by mkam last updated on 02/Jan/22

$$yessir$$

Answered by mr W last updated on 02/Jan/22

$${\left(\frac{dy}{dx}\right)}^3+x=y$$$$letu=y-x$$$$\frac{du}{dx}=\frac{dy}{dx}-1$$$${(\frac{du}{dx}+1)}^3=u$$$$\frac{du}{dx}=\sqrt[3]{u}-1$$$$\frac{du}{\sqrt[3]{u}-1}=dx$$$$\int \frac{du}{\sqrt[3]{u}-1}=\int dx$$$$lett=\sqrt[3]{u}-1$$$$u={(t+1)}^3$$$$du=3{(t+1)}^{2}dt$$$$\int \frac{3{(t+1)}^2}{t}dt=\int dx$$$$3\int (t+2+\frac{1}{t})dt=\int dx$$$$3(\frac{t^2}{2}+2t+\ln t)+C=x$$$$\frac{{(t+2)}^2}{2}+\ln t+C=\frac{x}{3}$$$$\frac{{(\sqrt[3]{u}+1)}^2}{2}+\ln (\sqrt[3]{u}-1)+C=\frac{x}{3}$$$$\Rightarrow \frac{{(\sqrt[3]{y-x}+1)}^2}{2}+\ln (\sqrt[3]{y-x}-1)+C=\frac{x}{3}$$

Commented by Kamel last updated on 02/Jan/22

$$ThisisaGaussdifferentialequation$$$$forexampley=1+xisasingularsolution.$$$$Ifweconsiderthatp=\frac{dy}{dx}wegetalinear$$$$differentialequationwithx,thenifwesolveitweget$$$$ageneralsolutionofequationwith$$$$parameterp=Const.$$

Commented by mr W last updated on 02/Jan/22

$$i{don}^{\prime }tunderstandthatmuch.what$$$$isthenthesolution?isanything$$$$wronginmysolution?$$

Commented by Kamel last updated on 02/Jan/22

$$No,{it}^{\prime }scorrect.$$

Commented by mr W last updated on 02/Jan/22

$$thanksforconfirmingsir!$$

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