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Question Number 26127 by sorour87 last updated on 20/Dec/17
y+2y^3 y^((1)) =(x+4yln (y))y^((1))
$${y}+\mathrm{2}{y}^{\mathrm{3}} {y}^{\left(\mathrm{1}\right)} =\left({x}+\mathrm{4}{y}\mathrm{ln}\:\left({y}\right)\right){y}^{\left(\mathrm{1}\right)} \\ $$
Answered by prakash jain last updated on 21/Dec/17
y+2y^3 (dy/dx)=(x+4yln y)(dy/dx) y+(2y^3 −x−4yln y)(dy/dx)=0 (i) ydx+(2y^3 −x−4yln y)dy=0 M(x,y)=y N(x,y)=(2y^3 −x−4yln y) (∂M/∂y)=1 (∂N/∂x)=−1 (∂M/∂y)≠(∂N/∂x) so equation is not exact. need to find u(x,y) such that (∂/∂y)(uM)=(∂/∂x)(uN) (ii) (((∂N/∂x)−(∂M/∂y))/M)=−(2/y) (function of y alone) so u is function of y alone u(y)=e^((∫ (((∂N/∂x)−(∂M/∂y))/M)dy)) =e^(∫−(2/y)dy) =e^(−2ln y) =(1/y^2 ) multiplying (i) by (1/y^2 ) (dx/y)−(((x+4yln y−2y^3 )/y^2 ))dy=0 now the equation is exact M=(1/y),N=−((x+4yln y−2y^3 )/y^2 ) (∂M/∂y)=−(1/y^2 ) so we F(x,y) such that ((∂F(x,y))/∂x)=M(x,y) ((∂F(x,y))/∂y)=N(x,y) F(x,y)=∫Mdx+g(y)=(x/y)+g(y) (∂F/∂y)=−(x/y^2 )+(dg/dy)=N(x,y)=−((x+4yln y−2y^3 )/y^2 ) (dg/dy)=−((4yln y−2y^3 )/y^2 ) (dg/dy)=2y−((4ln y)/y) g(y)=y^2 −2(ln y)^2 F(x,y)=(x/y)+g(y)=(x/y)+y^2 −2(ln y)^2 solution F(x,y)=c
$${y}+\mathrm{2}{y}^{\mathrm{3}} \frac{{dy}}{{dx}}=\left({x}+\mathrm{4}{y}\mathrm{ln}\:{y}\right)\frac{{dy}}{{dx}} \\ $$$${y}+\left(\mathrm{2}{y}^{\mathrm{3}} −{x}−\mathrm{4}{y}\mathrm{ln}\:{y}\right)\frac{{dy}}{{dx}}=\mathrm{0}\:\:\:\:\left({i}\right) \\ $$$${ydx}+\left(\mathrm{2}{y}^{\mathrm{3}} −{x}−\mathrm{4}{y}\mathrm{ln}\:{y}\right){dy}=\mathrm{0} \\ $$$${M}\left({x},{y}\right)={y} \\ $$$${N}\left({x},{y}\right)=\left(\mathrm{2}{y}^{\mathrm{3}} −{x}−\mathrm{4}{y}\mathrm{ln}\:{y}\right) \\ $$$$\frac{\partial{M}}{\partial{y}}=\mathrm{1} \\ $$$$\frac{\partial{N}}{\partial{x}}=−\mathrm{1} \\ $$$$\frac{\partial{M}}{\partial{y}}\neq\frac{\partial{N}}{\partial{x}}\:\mathrm{so}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{not}\:\mathrm{exact}. \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:{u}\left({x},{y}\right)\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{\partial}{\partial{y}}\left({uM}\right)=\frac{\partial}{\partial{x}}\left({uN}\right)\:\:\left({ii}\right) \\ $$$$\frac{\frac{\partial{N}}{\partial{x}}−\frac{\partial{M}}{\partial{y}}}{{M}}=−\frac{\mathrm{2}}{{y}}\:\left({function}\:{of}\:{y}\:{alone}\right) \\ $$$${so}\:\:{u}\:{is}\:{function}\:{of}\:{y}\:{alone} \\ $$$${u}\left({y}\right)={e}^{\left(\int\:\frac{\frac{\partial{N}}{\partial{x}}−\frac{\partial{M}}{\partial{y}}}{{M}}{dy}\right)} \\ $$$$={e}^{\int−\frac{\mathrm{2}}{{y}}{dy}} ={e}^{−\mathrm{2ln}\:{y}} =\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\mathrm{multiplying}\:\left(\mathrm{i}\right)\:\mathrm{by}\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\frac{{dx}}{{y}}−\left(\frac{{x}+\mathrm{4}{y}\mathrm{ln}\:{y}−\mathrm{2}{y}^{\mathrm{3}} }{{y}^{\mathrm{2}} }\right){dy}=\mathrm{0} \\ $$$${now}\:{the}\:{equation}\:{is}\:{exact} \\ $$$${M}=\frac{\mathrm{1}}{{y}},{N}=−\frac{{x}+\mathrm{4}{y}\mathrm{ln}\:{y}−\mathrm{2}{y}^{\mathrm{3}} }{{y}^{\mathrm{2}} } \\ $$$$\frac{\partial{M}}{\partial{y}}=−\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${so}\:{we}\:{F}\left({x},{y}\right)\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{\partial{F}\left({x},{y}\right)}{\partial{x}}={M}\left({x},{y}\right) \\ $$$$\frac{\partial{F}\left({x},{y}\right)}{\partial{y}}={N}\left({x},{y}\right) \\ $$$${F}\left({x},{y}\right)=\int{Mdx}+{g}\left({y}\right)=\frac{{x}}{{y}}+{g}\left({y}\right) \\ $$$$\frac{\partial{F}}{\partial{y}}=−\frac{{x}}{{y}^{\mathrm{2}} }+\frac{{dg}}{{dy}}={N}\left({x},{y}\right)=−\frac{{x}+\mathrm{4}{y}\mathrm{ln}\:{y}−\mathrm{2}{y}^{\mathrm{3}} }{{y}^{\mathrm{2}} } \\ $$$$\frac{{dg}}{{dy}}=−\frac{\mathrm{4}{y}\mathrm{ln}\:{y}−\mathrm{2}{y}^{\mathrm{3}} }{{y}^{\mathrm{2}} } \\ $$$$\frac{{dg}}{{dy}}=\mathrm{2}{y}−\frac{\mathrm{4ln}\:{y}}{{y}} \\ $$$${g}\left({y}\right)={y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ln}\:{y}\right)^{\mathrm{2}} \\ $$$${F}\left({x},{y}\right)=\frac{{x}}{{y}}+{g}\left({y}\right)=\frac{{x}}{{y}}+{y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ln}\:{y}\right)^{\mathrm{2}} \\ $$$${solution} \\ $$$${F}\left({x},{y}\right)={c} \\ $$
Commented by sorour87 last updated on 22/Dec/17
thank you
$$\boldsymbol{{thank}}\:\boldsymbol{{you}} \\ $$

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