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Question Number 97751 by bobhans last updated on 09/Jun/20

(y^2 −xy)dx + 2xy dy = 0

$$\left(\mathrm{y}^{\mathrm{2}} −\mathrm{xy}\right)\mathrm{dx}\:+\:\mathrm{2xy}\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$

Commented by bemath last updated on 09/Jun/20

⇔ 2xy dy = (xy−y^2 ) dx set y = zx ⇔ (dy/dx) = z + x (dz/dx) ⇔ z + x (dy/dx) = ((x^2 z−x^2 z^2 )/(2x^2 z)) z+ x (dy/dx) = ((z−z^2 )/(2z)) x (dy/dx) = ((z−3z^2 )/(2z)) ⇒((2z dz)/(z−3z^2 )) = (dx/x) ((2 dz)/(3z−1)) = −(dx/x) ⇒ (2/3)ln(3z−1)= −ln x +c (2/3)ln(((3y)/x)−1) = ln((C/x)) ln(((3y)/x)−1) = ln (√(((C/x))^3 )) ((3y)/x) = 1+ (√(((C/x))^3 )) y = (1/3)(x+x (√(((C/x))^3 )))

$$\Leftrightarrow\:\mathrm{2xy}\:\mathrm{dy}\:=\:\left(\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dx}\: \\ $$$$\mathrm{set}\:\mathrm{y}\:=\:\mathrm{zx}\:\Leftrightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}} \\ $$$$\Leftrightarrow\:{z}\:+\:{x}\:\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} {z}−{x}^{\mathrm{2}} {z}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} {z}} \\ $$$${z}+\:{x}\:\frac{{dy}}{{dx}}\:=\:\frac{{z}−{z}^{\mathrm{2}} }{\mathrm{2}{z}}\: \\ $$$${x}\:\frac{{dy}}{{dx}}\:=\:\frac{{z}−\mathrm{3}{z}^{\mathrm{2}} }{\mathrm{2}{z}}\:\Rightarrow\frac{\mathrm{2}{z}\:{dz}}{{z}−\mathrm{3}{z}^{\mathrm{2}} }\:=\:\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{2}\:{dz}}{\mathrm{3}{z}−\mathrm{1}}\:=\:−\frac{{dx}}{{x}}\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{3}{z}−\mathrm{1}\right)=\:−\mathrm{ln}\:{x}\:+{c} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\left(\frac{\mathrm{3y}}{{x}}−\mathrm{1}\right)\:=\:\mathrm{ln}\left(\frac{\mathrm{C}}{{x}}\right) \\ $$$$\mathrm{ln}\left(\frac{\mathrm{3y}}{{x}}−\mathrm{1}\right)\:=\:\mathrm{ln}\:\sqrt{\left(\frac{\mathrm{C}}{{x}}\right)^{\mathrm{3}} } \\ $$$$\frac{\mathrm{3}{y}}{{x}}\:=\:\mathrm{1}+\:\sqrt{\left(\frac{{C}}{{x}}\right)^{\mathrm{3}} } \\ $$$${y}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left({x}+{x}\:\sqrt{\left(\frac{{C}}{{x}}\right)^{\mathrm{3}} }\right)\: \\ $$

Commented by bobhans last updated on 09/Jun/20

waw.......

$$\mathrm{waw}....... \\ $$

Commented by john santu last updated on 10/Jun/20

yes..correct

$$\mathrm{yes}..\mathrm{correct}\: \\ $$

Answered by smridha last updated on 09/Jun/20

(∂/∂y)(y^2 −xy)=2y−x≠(∂/∂x)(2xy)=2y this is not exact differential so, (1/(2xy))[2y−x−2y]=−(1/(2y)) I.F=e^(∫−(dy/(2y))) =(1/(√y)) multi:by this of both sides of given eq^n (y^(3/2) −xy^(1/2) )dx+2xy^(1/2) dy=0 now ∫_((keeping y as constant)) (y^(3/2) −xy^(1/2) )dx =xy^(3/2) −(x^2 /2)y^(1/2) and ∫_((terms are donot containing x)) (0)dy=0 so solution: xy^(3/2) −(x^2 /2)y^(1/2) =c

$$\frac{\partial}{\partial{y}}\left(\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{xy}}\right)=\mathrm{2}\boldsymbol{{y}}−\boldsymbol{{x}}\neq\frac{\partial}{\partial\boldsymbol{{x}}}\left(\mathrm{2}\boldsymbol{{xy}}\right)=\mathrm{2}\boldsymbol{{y}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{exact}}\:\boldsymbol{{differential}} \\ $$$$\boldsymbol{{so}},\:\frac{\mathrm{1}}{\mathrm{2}{xy}}\left[\mathrm{2}{y}−\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{y}}\right]=−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{y}}} \\ $$$$\boldsymbol{{I}}.{F}={e}^{\int−\frac{{dy}}{\mathrm{2}\boldsymbol{{y}}}} =\frac{\mathrm{1}}{\sqrt{\boldsymbol{{y}}}}\:\:\:\boldsymbol{{multi}}:\boldsymbol{{by}}\:\boldsymbol{{this}}\:\boldsymbol{{of}}\:\boldsymbol{{both}} \\ $$$$\boldsymbol{{sides}}\:\boldsymbol{{of}}\:\boldsymbol{{given}}\:\boldsymbol{{eq}}^{\boldsymbol{{n}}} \\ $$$$\left(\boldsymbol{{y}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\boldsymbol{{xy}}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)\boldsymbol{{dx}}+\mathrm{2}\boldsymbol{{xy}}^{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dy}}=\mathrm{0} \\ $$$$\boldsymbol{{now}}\:\int_{\left({keeping}\:\boldsymbol{{y}}\:\boldsymbol{{as}}\:\boldsymbol{{constant}}\right)} \left(\boldsymbol{{y}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\boldsymbol{{xy}}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)\boldsymbol{{dx}} \\ $$$$=\boldsymbol{{xy}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}}\boldsymbol{{y}}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\boldsymbol{{and}}\:\int_{\left(\boldsymbol{{terms}}\:\boldsymbol{{are}}\:\boldsymbol{{donot}}\:\boldsymbol{{containing}}\:\boldsymbol{{x}}\right)} \left(\mathrm{0}\right)\boldsymbol{{dy}}=\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{solution}}: \\ $$$$\:\:\:\:{x}\boldsymbol{{y}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{2}}\boldsymbol{{y}}^{\frac{\mathrm{1}}{\mathrm{2}}} =\boldsymbol{{c}} \\ $$

Commented by mr W last updated on 09/Jun/20

please check sir: with xy^(3/2) −(x^2 /2)y^(1/2) =c we get: y^(3/2) +(3/2)xy^(1/2) (dy/dx)−xy^(1/2) −(x^2 /4)y^(−(1/2)) (dy/dx)=0 4y(y−x)+x(6y−x)(dy/dx)=0 ⇒4y(y−x)dx+x(6y−x)dy=0 this is not the original d. eqn.

$${please}\:{check}\:{sir}: \\ $$$${with}\:{xy}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{y}^{\frac{\mathrm{1}}{\mathrm{2}}} ={c}\:{we}\:{get}: \\ $$$${y}^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{3}}{\mathrm{2}}{xy}^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dy}}{{dx}}−{xy}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{x}^{\mathrm{2}} }{\mathrm{4}}{y}^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{4}{y}\left({y}−{x}\right)+{x}\left(\mathrm{6}{y}−{x}\right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{y}\left({y}−{x}\right){dx}+{x}\left(\mathrm{6}{y}−{x}\right){dy}=\mathrm{0} \\ $$$${this}\:{is}\:{not}\:{the}\:{original}\:{d}.\:{eqn}. \\ $$

Commented by smridha last updated on 09/Jun/20

I donot find any mistake.. if you can then show me....

$$\boldsymbol{{I}}\:{do}\boldsymbol{{not}}\:\boldsymbol{{find}}\:\boldsymbol{{any}}\:\boldsymbol{{mistake}}.. \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{then}}\:\boldsymbol{{show}}\:\boldsymbol{{me}}.... \\ $$

Commented by prakash jain last updated on 10/Jun/20

(y^2 −xy)dx+2xydy=0 Mdx+Ndy=0 M_y =2y−x,N_x =2y ((M_y −N_x )/N)=((−x)/(2xy))=−(1/(2y)) I think the mistake is in calculating I.F. It is function of y. It should be function of x to use as integrating factor. (y^(3/2) −xy^(1/2) )dx+2xy^(1/2) dy=0 So the above equation is still not exact.

$$\left({y}^{\mathrm{2}} −{xy}\right){dx}+\mathrm{2}{xydy}=\mathrm{0} \\ $$$${Mdx}+{Ndy}=\mathrm{0} \\ $$$${M}_{{y}} =\mathrm{2}{y}−{x},{N}_{{x}} =\mathrm{2}{y} \\ $$$$\frac{{M}_{{y}} −{N}_{{x}} }{{N}}=\frac{−{x}}{\mathrm{2}{xy}}=−\frac{\mathrm{1}}{\mathrm{2}{y}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{is}\:\mathrm{in} \\ $$$$\mathrm{calculating}\:\mathrm{I}.\mathrm{F}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{function}\:\mathrm{of}\:\boldsymbol{{y}}.\:\mathrm{It}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{function}\:\mathrm{of}\:\boldsymbol{{x}}\:\mathrm{to}\:\mathrm{use}\:\mathrm{as}\:\mathrm{integrating} \\ $$$$\mathrm{factor}. \\ $$$$\left(\boldsymbol{{y}}^{\frac{\mathrm{3}}{\mathrm{2}}} −\boldsymbol{{xy}}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)\boldsymbol{{dx}}+\mathrm{2}\boldsymbol{{xy}}^{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dy}}=\mathrm{0} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{above}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{still}\:\mathrm{not}\:\mathrm{exact}. \\ $$

Commented by smridha last updated on 10/Jun/20

yeah my second solution tells the truth ....but i am not yet understand why the IF is different for this two mehods.....two method sshould be analogus to each other....

$${ye}\boldsymbol{{ah}}\:\boldsymbol{{my}}\:\boldsymbol{{second}}\:\boldsymbol{{solution}}\:\boldsymbol{{tells}} \\ $$$$\boldsymbol{{th}}{e}\:\boldsymbol{{truth}}\:....\boldsymbol{{but}}\:\boldsymbol{{i}}\:\boldsymbol{{am}}\:\boldsymbol{{not}}\:\boldsymbol{{yet}} \\ $$$$\boldsymbol{{understand}}\:\boldsymbol{{why}}\:\boldsymbol{{the}}\:\boldsymbol{{I}}{F}\:\boldsymbol{{is}}\:\boldsymbol{{different}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{this}}\:\boldsymbol{{two}}\:\boldsymbol{{mehods}}.....\boldsymbol{{two}} \\ $$$$\boldsymbol{{method}}\:{s}\boldsymbol{{should}}\:\boldsymbol{{be}}\:\boldsymbol{{analogus}} \\ $$$$\boldsymbol{{t}}{o}\:\boldsymbol{{each}}\:\boldsymbol{{other}}.... \\ $$$$ \\ $$

Commented by prakash jain last updated on 10/Jun/20

((M_y −N_x )/N) should not have y and should be a fuction of only x. So there is a mistake in method used for computing I.F in the answer above

$$\frac{{M}_{{y}} −{N}_{{x}} }{{N}}\:\mathrm{should}\:\mathrm{not}\:\mathrm{have}\:{y}\:\mathrm{and} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{a}\:\mathrm{fuction}\:\mathrm{of}\:\mathrm{only}\:{x}. \\ $$$$\mathrm{So}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{method}\: \\ $$$$\mathrm{used}\:\mathrm{for}\:\mathrm{computing}\:\mathrm{I}.\mathrm{F}\:\mathrm{in}\:\mathrm{the}\:\mathrm{answer} \\ $$$$\mathrm{above} \\ $$

Answered by smridha last updated on 10/Jun/20

(y^2 −xy)dx+2xydy=0 (dy/dx)+(y/(2x))=(1/2) so IF=e^(∫(dx/(2x))) =(√x) mult: and integrating by both sides with this we get ∫d(y(√x))=(1/2)∫(√x)dx so the solution is y(√x)=(1/3)x^(3/2) +c now this is perfect....but still I have confution why my previous method didnot work...probably I did silly mistake anywhere....

$$\left(\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{xy}}\right)\boldsymbol{{dx}}+\mathrm{2}\boldsymbol{{xydy}}=\mathrm{0} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\frac{\boldsymbol{{y}}}{\mathrm{2}\boldsymbol{{x}}}=\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{{so}}\:\boldsymbol{{IF}}=\boldsymbol{{e}}^{\int\frac{\boldsymbol{{dx}}}{\mathrm{2}\boldsymbol{{x}}}} =\sqrt{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{mult}}:\:\:\boldsymbol{{and}}\:\boldsymbol{{integrating}}\:\boldsymbol{{by}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\:\boldsymbol{{with}}\:\boldsymbol{{this}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\int\boldsymbol{{d}}\left(\boldsymbol{{y}}\sqrt{\boldsymbol{{x}}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}}{dx} \\ $$$${so}\:{the}\:\boldsymbol{{solution}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{y}}\sqrt{\boldsymbol{{x}}}=\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} +\boldsymbol{{c}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{perfect}}....\boldsymbol{{but}}\:\boldsymbol{{still}}\:\boldsymbol{{I}} \\ $$$$\boldsymbol{{have}}\:\boldsymbol{{confution}}\:\boldsymbol{{why}}\:\boldsymbol{{my}}\:\boldsymbol{{previous}} \\ $$$$\boldsymbol{{method}}\:\boldsymbol{{didnot}}\:\boldsymbol{{work}}...\boldsymbol{{probably}} \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{did}}\:\boldsymbol{{silly}}\:\boldsymbol{{mistake}}\:\boldsymbol{{anywhere}}.... \\ $$$$ \\ $$

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