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Question Number 92040 by mhmd last updated on 04/May/20
y^(′′) +(y′)^2 +y=0 y(0)=−(1/2) , y′(0)=−1 help me sir
$${y}^{''} +\left({y}'\right)^{\mathrm{2}} +{y}=\mathrm{0}\:\:\:{y}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:,\:{y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$${help}\:{me}\:{sir}\: \\ $$
Answered by mr W last updated on 04/May/20
let u=y′=(dy/dx) y′′=(du/dx)=(du/dy)×(dy/dx)=u(du/dy) u(du/dy)+u^2 +y=0 (1/2)((d(u^2 ))/dy)+u^2 +y=0 let U=u^2 (1/2)(dU/dy)+U+y=0 (dU/dy)+2U=−2y U=((−∫2ye^(∫2dy) dy+C)/e^(∫2fy) ) U=((−∫2ye^(2y) dy+C)/e^(2y) ) U=((−ye^(2y) +∫e^(2y) dy+C)/e^(2y) ) U=((−ye^(2y) +(1/2)e^(2y) +C)/e^(2y) ) U=((1−2y+C_1 e^(−2y) )/2)=u^2 =((dy/dx))^2 (dy/dx)=±((√(1−2y+C_1 e^(−2y) ))/( (√2))) −1=±((√(2+C_1 e))/( (√2))) ⇒C_1 =0 (dy/( (√(1−2y))))=−(dx/( (√2))) x=−(√2)∫(dy/( (√(1−2y))))+C_2 x=(√(2(1−2y)))+C_2 0=(√(2(1+1)))+C_2 ⇒C_2 =−2 ⇒x=(√(2(1−2y)))−2 or ⇒y=(1/2)−(1/4)(x+2)^2
$${let}\:{u}={y}'=\frac{{dy}}{{dx}} \\ $$$${y}''=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$${u}\frac{{du}}{{dy}}+{u}^{\mathrm{2}} +{y}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\frac{{d}\left({u}^{\mathrm{2}} \right)}{{dy}}+{u}^{\mathrm{2}} +{y}=\mathrm{0} \\ $$$${let}\:{U}={u}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\frac{{dU}}{{dy}}+{U}+{y}=\mathrm{0} \\ $$$$\frac{{dU}}{{dy}}+\mathrm{2}{U}=−\mathrm{2}{y} \\ $$$${U}=\frac{−\int\mathrm{2}{ye}^{\int\mathrm{2}{dy}} {dy}+{C}}{{e}^{\int\mathrm{2}{fy}} } \\ $$$${U}=\frac{−\int\mathrm{2}{ye}^{\mathrm{2}{y}} {dy}+{C}}{{e}^{\mathrm{2}{y}} } \\ $$$${U}=\frac{−{ye}^{\mathrm{2}{y}} +\int{e}^{\mathrm{2}{y}} {dy}+{C}}{{e}^{\mathrm{2}{y}} } \\ $$$${U}=\frac{−{ye}^{\mathrm{2}{y}} +\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{y}} +{C}}{{e}^{\mathrm{2}{y}} } \\ $$$${U}=\frac{\mathrm{1}−\mathrm{2}{y}+{C}_{\mathrm{1}} {e}^{−\mathrm{2}{y}} }{\mathrm{2}}={u}^{\mathrm{2}} =\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\pm\frac{\sqrt{\mathrm{1}−\mathrm{2}{y}+{C}_{\mathrm{1}} {e}^{−\mathrm{2}{y}} }}{\:\sqrt{\mathrm{2}}} \\ $$$$−\mathrm{1}=\pm\frac{\sqrt{\mathrm{2}+{C}_{\mathrm{1}} {e}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{C}_{\mathrm{1}} =\mathrm{0} \\ $$$$\frac{{dy}}{\:\sqrt{\mathrm{1}−\mathrm{2}{y}}}=−\frac{{dx}}{\:\sqrt{\mathrm{2}}} \\ $$$${x}=−\sqrt{\mathrm{2}}\int\frac{{dy}}{\:\sqrt{\mathrm{1}−\mathrm{2}{y}}}+{C}_{\mathrm{2}} \\ $$$${x}=\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{y}\right)}+{C}_{\mathrm{2}} \\ $$$$\mathrm{0}=\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{1}\right)}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{C}_{\mathrm{2}} =−\mathrm{2} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{y}\right)}−\mathrm{2} \\ $$$${or} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{2}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/May/20
check: y(0)=(1/2)−1=−(1/2) y′=−((x+2)/2)=−1−(x/2) y′(0)=−1 y′′=−(1/2) y′′+(y′)^2 +y=−(1/2)+(((x+2)^2 )/4)+(1/2)−(((x+2)^2 )/4)=0
$${check}: \\ $$$${y}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}'=−\frac{{x}+\mathrm{2}}{\mathrm{2}}=−\mathrm{1}−\frac{{x}}{\mathrm{2}} \\ $$$${y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$${y}''=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}''+\left({y}'\right)^{\mathrm{2}} +{y}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$
Commented by niroj last updated on 04/May/20
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