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Question Number 121028 by bramlexs22 last updated on 05/Nov/20
Answered by liberty last updated on 05/Nov/20
![no. this not separable but it is a homogenous diff equation. ⇒ (dy/dx) = ((x^4 +2y^4 )/(xy^3 )) ; [ put y = zx ] ⇒z + x (dz/dx) = ((x^4 (1+2z^4 ))/(z^3 x^4 ))=((1+2z^4 )/z^3 ) ⇒z + x (dz/dx) = (1/z^3 )+2z ; x(dz/dx) = ((1+z^4 )/z^3 ) ⇒ (z^3 /(1+z^4 )) dz = (dx/x) ; ∫ ((d(1+z^4 ))/(1+z^4 )) = 4∫(dx/x) ⇒ln (1+z^4 )=ln (Cx^4 ) ⇒1+((y/x))^4 = Cx^4 ⇒x^4 +y^4 =Cx^8](Q121038.png)
$$\mathrm{no}.\:\mathrm{this}\:\mathrm{not}\:\mathrm{separable}\:\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{homogenous} \\ $$$$\mathrm{diff}\:\mathrm{equation}. \\ $$$$\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{2y}^{\mathrm{4}} }{\mathrm{xy}^{\mathrm{3}} }\:\:;\:\left[\:\mathrm{put}\:\mathrm{y}\:=\:\mathrm{zx}\:\right] \\ $$$$\Rightarrow\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}+\mathrm{2z}^{\mathrm{4}} \right)}{\mathrm{z}^{\mathrm{3}} \mathrm{x}^{\mathrm{4}} }=\frac{\mathrm{1}+\mathrm{2z}^{\mathrm{4}} }{\mathrm{z}^{\mathrm{3}} } \\ $$$$\Rightarrow\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{3}} }+\mathrm{2z}\:;\:\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}+\mathrm{z}^{\mathrm{4}} }{\mathrm{z}^{\mathrm{3}} } \\ $$$$\Rightarrow\:\frac{\mathrm{z}^{\mathrm{3}} }{\mathrm{1}+\mathrm{z}^{\mathrm{4}} }\:\mathrm{dz}\:=\:\frac{\mathrm{dx}}{\mathrm{x}}\:;\:\int\:\frac{\mathrm{d}\left(\mathrm{1}+\mathrm{z}^{\mathrm{4}} \right)}{\mathrm{1}+\mathrm{z}^{\mathrm{4}} }\:=\:\mathrm{4}\int\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{ln}\:\left(\mathrm{1}+\mathrm{z}^{\mathrm{4}} \right)=\mathrm{ln}\:\left(\mathrm{Cx}^{\mathrm{4}} \right)\: \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{4}} =\:\mathrm{Cx}^{\mathrm{4}} \:\Rightarrow\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} =\mathrm{Cx}^{\mathrm{8}} \\ $$