Menu Close

Find-general-solutions-the-following-differential-equations-a-d-2-y-dx-2-dy-dx-6y-b-3x-2-y-2-dx-x-2-y-2-dy-0-

Tinku Tara 0 Comments
Pin




Question Number 175464 by nadovic last updated on 31/Aug/22
Find general solutions the following differential equations (a) (d^2 y/dx^2 ) + (dy/dx) = 6y (b) (3x^2 +y^2 )dx + (x^2 +y^2 )dy = 0
$$\:\mathrm{Find}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{differential}\:\mathrm{equations} \\ $$$$\left({a}\right)\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}\:=\:\mathrm{6}{y} \\ $$$$\left(\mathrm{b}\right)\:\:\:\left(\mathrm{3}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dx}\:+\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by floor(10²Eta[1]) last updated on 31/Aug/22
y′′+y′−6y=0 characteristic equation: D^2 +D−6=0⇒D=−3, 2 ⇒y=Ae^(−3x) +Be^(2x)
$$\mathrm{y}''+\mathrm{y}'−\mathrm{6y}=\mathrm{0} \\ $$$$\mathrm{characteristic}\:\mathrm{equation}: \\ $$$$\mathrm{D}^{\mathrm{2}} +\mathrm{D}−\mathrm{6}=\mathrm{0}\Rightarrow\mathrm{D}=−\mathrm{3},\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{y}=\mathrm{Ae}^{−\mathrm{3x}} +\mathrm{Be}^{\mathrm{2x}} \\ $$
Answered by floor(10²Eta[1]) last updated on 31/Aug/22
y(x)=xv(x) y′(x)=v(x)+xv′(x)⇒(dy/dx)=v+x(dv/dx) dy=vdx+xdv (3x^2 +x^2 v^2 )dx+(x^2 +x^2 v^2 )(vdx+xdv) (3x^2 +x^2 v^2 +x^2 v+x^2 v^3 )dx+(x^3 +x^3 v^2 )dv=0 (3+v^2 +v+v^3 )dx+(x+xv^2 )dv=0 (dv/dx)=−((v^3 +v^2 +v+3)/(x(v^2 +1))) −((v^2 +1)/(v^3 +v^2 +v+3))dv=(dx/x) −∫((v^2 +1)/(v^3 +v^2 +v+3))dv=∫(dx/x)
$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{xv}\left(\mathrm{x}\right) \\ $$$$\mathrm{y}'\left(\mathrm{x}\right)=\mathrm{v}\left(\mathrm{x}\right)+\mathrm{xv}'\left(\mathrm{x}\right)\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\mathrm{dy}=\mathrm{vdx}+\mathrm{xdv} \\ $$$$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} \right)\left(\mathrm{vdx}+\mathrm{xdv}\right) \\ $$$$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{v}+\mathrm{x}^{\mathrm{2}} \mathrm{v}^{\mathrm{3}} \right)\mathrm{dx}+\left(\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{3}} \mathrm{v}^{\mathrm{2}} \right)\mathrm{dv}=\mathrm{0} \\ $$$$\left(\mathrm{3}+\mathrm{v}^{\mathrm{2}} +\mathrm{v}+\mathrm{v}^{\mathrm{3}} \right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{xv}^{\mathrm{2}} \right)\mathrm{dv}=\mathrm{0} \\ $$$$\frac{\mathrm{dv}}{\mathrm{dx}}=−\frac{\mathrm{v}^{\mathrm{3}} +\mathrm{v}^{\mathrm{2}} +\mathrm{v}+\mathrm{3}}{\mathrm{x}\left(\mathrm{v}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$−\frac{\mathrm{v}^{\mathrm{2}} +\mathrm{1}}{\mathrm{v}^{\mathrm{3}} +\mathrm{v}^{\mathrm{2}} +\mathrm{v}+\mathrm{3}}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$−\int\frac{\mathrm{v}^{\mathrm{2}} +\mathrm{1}}{\mathrm{v}^{\mathrm{3}} +\mathrm{v}^{\mathrm{2}} +\mathrm{v}+\mathrm{3}}\mathrm{dv}=\int\frac{\mathrm{dx}}{\mathrm{x}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *