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Question Number 118376 by bramlexs22 last updated on 17/Oct/20
$${If}\:{a}\:{curve}\:{y}\:=\:{f}\left({x}\right)\:{passing}\:{through} \\ $$$${the}\:{point}\:\left(\mathrm{1},\mathrm{2}\right)\:{is}\:{the}\:{solution} \\ $$$${of}\:{differential}\:{equation} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:{dy}\:=\:\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dx}\:,\:{then}\:{the}\: \\ $$$${value}\:{of}\:{f}\left(\mathrm{2}\right)\:{is}\:{equal}\:{to}? \\ $$
Answered by benjo_mathlover last updated on 17/Oct/20
![solving for differential equation (dy/dx) = ((2xy+y^2 )/(2x^2 )). [ set y = zx ] ⇒ x (dz/dx) + z = ((2x(zx)+z^2 x^2 )/(2x^2 )) ⇒ x (dz/dx) + z = ((2z+z^2 )/2) ⇒x(dz/dx) = (1/2)z^2 ; (dz/z^2 ) = (1/2) (dx/x) ⇒−(1/z) = (1/2)ln (x)+c ; or ⇒(x/y) = −(1/2)ln (x)−c ; substitute point(1,2) ⇒(1/2)=−(1/2)ln (1)−c ; c = −(1/2) thus (x/y) = −(1/2)ln (x)+(1/2) ⇒((2x)/y)= 1−ln (x) or y = ((2x)/(1−ln (x))) therefore f(2) = (4/(1−ln (2)))](Q118377.png)
$${solving}\:{for}\:{differential}\:{equation} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{xy}+{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }.\:\left[\:{set}\:{y}\:=\:{zx}\:\right]\: \\ $$$$\Rightarrow\:{x}\:\frac{{dz}}{{dx}}\:+\:{z}\:=\:\frac{\mathrm{2}{x}\left({zx}\right)+{z}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{x}\:\frac{{dz}}{{dx}}\:+\:{z}\:=\:\frac{\mathrm{2}{z}+{z}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{x}\frac{{dz}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{z}^{\mathrm{2}} \:;\:\frac{{dz}}{{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{dx}}{{x}} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{{z}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}\right)+{c}\:;\:{or}\: \\ $$$$\Rightarrow\frac{{x}}{{y}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}\right)−{c}\:;\:{substitute}\:{point}\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}\right)−{c}\:;\:{c}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${thus}\:\frac{{x}}{{y}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{2}{x}}{{y}}=\:\mathrm{1}−\mathrm{ln}\:\left({x}\right)\:{or}\:{y}\:=\:\frac{\mathrm{2}{x}}{\mathrm{1}−\mathrm{ln}\:\left({x}\right)} \\ $$$${therefore}\:{f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{4}}{\mathrm{1}−\mathrm{ln}\:\left(\mathrm{2}\right)} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\mathrm{2}{x}^{\mathrm{2}} {dy}−\mathrm{2}{xydx}={y}^{\mathrm{2}} {dx} \\ $$$$−\mathrm{2}{x}\left(\frac{{ydx}−{xdy}}{{y}^{\mathrm{2}} }\right)={dx} \\ $$$$−\mathrm{2}{d}\left(\frac{{x}}{{y}}\right)=\frac{{dx}}{{x}} \\ $$$$−\mathrm{2}\left(\frac{{x}}{{y}}\right)={lnx}+{lnc} \\ $$$${ln}\left({xc}\right)=\frac{−\mathrm{2}{x}}{{y}} \\ $$$${xc}={e}^{−\frac{\mathrm{2}{x}}{{y}}} \\ $$$$\mathrm{1}×{c}={e}^{−\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}} \rightarrow\boldsymbol{{c}}=\boldsymbol{{e}}^{−\mathrm{1}} \\ $$$$\boldsymbol{{xe}}^{−\mathrm{1}} =\boldsymbol{{e}}^{−\frac{\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{y}}}} \\ $$$$\left(\boldsymbol{{lnx}}\right)+\left(−\mathrm{1}\right)=\frac{−\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{y}}} \\ $$$$\boldsymbol{{y}}=\frac{−\mathrm{2}\boldsymbol{{x}}}{−\mathrm{1}+\boldsymbol{{lnx}}}\rightarrow{f}\left(\mathrm{2}\right)=\frac{−\mathrm{4}}{−\mathrm{1}+\boldsymbol{{ln}}\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{1}−\boldsymbol{{ln}}\mathrm{2}} \\ $$