Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 67899 by ramirez105 last updated on 01/Sep/19

homogenous differential equation. (2xy+y^2 )dr−2x^2 dy=0 y=e x=e

$${homogenous}\:{differential}\:{equation}. \\ $$$$ \\ $$$$\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dr}−\mathrm{2}{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$${y}={e} \\ $$$${x}={e} \\ $$

Commented by AnJan_Math_Max last updated on 02/Sep/19

First Of All, The Question May Be:-(2xy+y^2 )dx−2x^(2 ) dy=0

$$\boldsymbol{{First}}\:\boldsymbol{{Of}}\:\boldsymbol{{All}},\:\boldsymbol{{The}}\:\boldsymbol{{Question}}\:\boldsymbol{{May}}\:\boldsymbol{{Be}}:-\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dx}−\mathrm{2}{x}^{\mathrm{2}\:} {dy}=\mathrm{0}\: \\ $$

Commented by mathmax by abdo last updated on 02/Sep/19

(2xy+y^2 )dx−2x^2 dy =0 ⇒2x^2 dy =(2xy+y^2 )dx ⇒ (dy/(y^2 +2xy)) =(dx/(2x^2 )) ⇒∫ (dy/(y^2 +2xy)) =−(x/2) +c but ∫ (dy/(y^2 +2xy)) =∫ (dy/(y(2x+y))) =(1/(2x))∫ ((1/y)−(1/(2x+y)))=(1/(2x)){ln∣y∣−(1/2)ln(2x+y)} =(1/(2x))ln∣(y/(√(2x+y)))∣ ⇒(1/(2x))ln∣(y/(√(2x+y)))∣=−(x/2) +c ⇒ ln∣(y/(√(2x+y)))∣ =−1+2cx ⇒(y/(√(2x+y))) =e^(−1+2cx) ⇒ (y^2 /(2x+y)) =e^(−2+4cx) =λ(x) ⇒y^2 =λ(x)(2x+y) ⇒ y^2 −2xλ(x) +λ(x)y =0 ⇒y^2 +λ(x)y−2xλ(x)=0 Δ =λ^2 (x)+8xλ(x) ⇒y_1 =((−λ(x)+(√(λ^2 (x)+8xλ(x))))/2) and y_2 =((−λ(x)−(√(λ^2 (x)+8xλ(x))))/2) with λ(x) =e^(−2+4cx)

$$\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dx}−\mathrm{2}{x}^{\mathrm{2}} {dy}\:=\mathrm{0}\:\Rightarrow\mathrm{2}{x}^{\mathrm{2}} {dy}\:=\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dx}\:\Rightarrow \\ $$$$\frac{{dy}}{{y}^{\mathrm{2}} \:+\mathrm{2}{xy}}\:=\frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow\int\:\:\frac{{dy}}{{y}^{\mathrm{2}} \:+\mathrm{2}{xy}}\:=−\frac{{x}}{\mathrm{2}}\:+{c}\:\:{but} \\ $$$$\int\:\:\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{2}{xy}}\:=\int\:\:\frac{{dy}}{{y}\left(\mathrm{2}{x}+{y}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{x}}\int\:\:\left(\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{\mathrm{2}{x}+{y}}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}\left\{{ln}\mid{y}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}{x}+{y}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{x}}{ln}\mid\frac{{y}}{\sqrt{\mathrm{2}{x}+{y}}}\mid\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}{ln}\mid\frac{{y}}{\sqrt{\mathrm{2}{x}+{y}}}\mid=−\frac{{x}}{\mathrm{2}}\:+{c}\:\Rightarrow \\ $$$${ln}\mid\frac{{y}}{\sqrt{\mathrm{2}{x}+{y}}}\mid\:=−\mathrm{1}+\mathrm{2}{cx}\:\Rightarrow\frac{{y}}{\sqrt{\mathrm{2}{x}+{y}}}\:={e}^{−\mathrm{1}+\mathrm{2}{cx}} \:\Rightarrow \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}{x}+{y}}\:={e}^{−\mathrm{2}+\mathrm{4}{cx}} =\lambda\left({x}\right)\:\Rightarrow{y}^{\mathrm{2}} =\lambda\left({x}\right)\left(\mathrm{2}{x}+{y}\right)\:\Rightarrow \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{x}\lambda\left({x}\right)\:+\lambda\left({x}\right){y}\:=\mathrm{0}\:\Rightarrow{y}^{\mathrm{2}} +\lambda\left({x}\right){y}−\mathrm{2}{x}\lambda\left({x}\right)=\mathrm{0} \\ $$$$\Delta\:=\lambda^{\mathrm{2}} \left({x}\right)+\mathrm{8}{x}\lambda\left({x}\right)\:\:\Rightarrow{y}_{\mathrm{1}} =\frac{−\lambda\left({x}\right)+\sqrt{\lambda^{\mathrm{2}} \left({x}\right)+\mathrm{8}{x}\lambda\left({x}\right)}}{\mathrm{2}} \\ $$$${and}\:{y}_{\mathrm{2}} =\frac{−\lambda\left({x}\right)−\sqrt{\lambda^{\mathrm{2}} \left({x}\right)+\mathrm{8}{x}\lambda\left({x}\right)}}{\mathrm{2}} \\ $$$${with}\:\lambda\left({x}\right)\:={e}^{−\mathrm{2}+\mathrm{4}{cx}} \\ $$

Answered by AnJan_Math_Max last updated on 02/Sep/19

(2xy+y^2 )−2x^2 (dy/dx)=0 ⇒(dy/dx)=((2xy+y^2 )/(2x^2 )) This is a homogeneous differential equation. Putting y=vx and (dy/dx)=v+x(dv/dx) ,it reduces to v+x(dv/dx)=((2v+v^2 )/2) ⇒2v+2x(dv/dx)=2v+v^2 ⇒2x(dv/dx)=v^2 ⇒(2/v^2 )dv=(1/x)dx ⇒2∫(1/v^2 )dv=∫(1/x)dx ⇒−(2/v)=log ∣x∣+C ⇒−((2x)/y)=log ∣x∣+C ...(i) It is given that y=e,x=e. Putting x=e,y=e in (i),We get −2=1+C ⇒C=−3 ∴From (i): −((2x)/y)=log ∣x∣−3 ⇒y=((2x)/(3−log ∣x∣))

$$\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right)−\mathrm{2}{x}^{\mathrm{2}} \frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{2}{xy}+{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\boldsymbol{{This}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{homogeneous}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}}. \\ $$$$\boldsymbol{{Putting}}\:{y}={vx}\:\boldsymbol{{and}}\:\frac{{dy}}{{dx}}={v}+{x}\frac{{dv}}{{dx}}\:,\boldsymbol{{it}}\:\boldsymbol{{reduces}}\:\boldsymbol{{to}} \\ $$$${v}+{x}\frac{{dv}}{{dx}}=\frac{\mathrm{2}{v}+{v}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{v}+\mathrm{2}{x}\frac{{dv}}{{dx}}=\mathrm{2}{v}+{v}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{x}\frac{{dv}}{{dx}}={v}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{v}^{\mathrm{2}} }{dv}=\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\Rightarrow\mathrm{2}\int\frac{\mathrm{1}}{{v}^{\mathrm{2}} }{dv}=\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\Rightarrow−\frac{\mathrm{2}}{{v}}=\mathrm{log}\:\mid{x}\mid+{C}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\Rightarrow−\frac{\mathrm{2}{x}}{{y}}=\mathrm{log}\:\mid{x}\mid+{C}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$$$\boldsymbol{{It}}\:\boldsymbol{{is}}\:\boldsymbol{{given}}\:\boldsymbol{{that}}\:{y}={e},{x}={e}. \\ $$$$\boldsymbol{{Putting}}\:{x}={e},{y}={e}\:\boldsymbol{{in}}\:\left(\boldsymbol{\mathrm{i}}\right),\boldsymbol{{We}}\:\boldsymbol{{get}} \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}=\mathrm{1}+\mathrm{C} \\ $$$$\Rightarrow{C}=−\mathrm{3} \\ $$$$\therefore\boldsymbol{{From}}\:\left(\boldsymbol{\mathrm{i}}\right): \\ $$$$−\frac{\mathrm{2}{x}}{{y}}=\mathrm{log}\:\mid{x}\mid−\mathrm{3} \\ $$$$\Rightarrow\boldsymbol{{y}}=\frac{\mathrm{2}\boldsymbol{{x}}}{\mathrm{3}−\mathrm{log}\:\mid\boldsymbol{{x}}\mid} \\ $$

Terms of Service

Privacy Policy

Contact: [email protected]