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Question Number 6661 by Tawakalitu. last updated on 09/Jul/16

Solve this equation by reducing it from non homogeneous  equation to homogeneous equation  (dy/dx) = ((2(x + y − 1))/(3x + y + 1))    Please help with this one too.  i was trying your approah but not the same    thanks for your help.

$${Solve}\:{this}\:{equation}\:{by}\:{reducing}\:{it}\:{from}\:{non}\:{homogeneous} \\ $$$${equation}\:{to}\:{homogeneous}\:{equation} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}\left({x}\:+\:{y}\:−\:\mathrm{1}\right)}{\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{1}} \\ $$$$ \\ $$$${Please}\:{help}\:{with}\:{this}\:{one}\:{too}. \\ $$$${i}\:{was}\:{trying}\:{your}\:{approah}\:{but}\:{not}\:{the}\:{same} \\ $$$$ \\ $$$${thanks}\:{for}\:{your}\:{help}. \\ $$

Commented by prakash jain last updated on 11/Jul/16

u=x+a  v=y+b  2a+2b−1=0  3a+b+1=0⇒6a+2b+2=0  4a+3=0⇒a=−3/4  b=5/4  u=x−3/4  v=y+5/4  (dv/du)=((2u+2v)/(3u+v))  Solve by substituting u/v=t

$${u}={x}+{a} \\ $$$${v}={y}+{b} \\ $$$$\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}{a}+{b}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{6}{a}+\mathrm{2}{b}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{a}+\mathrm{3}=\mathrm{0}\Rightarrow{a}=−\mathrm{3}/\mathrm{4} \\ $$$${b}=\mathrm{5}/\mathrm{4} \\ $$$${u}={x}−\mathrm{3}/\mathrm{4} \\ $$$${v}={y}+\mathrm{5}/\mathrm{4} \\ $$$$\frac{{dv}}{{du}}=\frac{\mathrm{2}{u}+\mathrm{2}{v}}{\mathrm{3}{u}+{v}} \\ $$$$\mathrm{Solve}\:\mathrm{by}\:\mathrm{substituting}\:{u}/{v}={t} \\ $$

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