Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 6651 by Tawakalitu. last updated on 08/Jul/16

Solve this equation by reducing it from non homogeneous equation to homogeneous equation (dy/dx) = ((x + y + 3)/(x − y −5))

$${Solve}\:{this}\:{equation}\:{by}\:{reducing}\:{it}\:{from}\:{non}\:{homogeneous} \\ $$$${equation}\:{to}\:{homogeneous}\:{equation}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}\:+\:{y}\:+\:\mathrm{3}}{{x}\:−\:{y}\:−\mathrm{5}} \\ $$

Commented by prakash jain last updated on 09/Jul/16

Substitue x=u+1 and y=v−4 (dv/du)=((u+v)/(u−v))

$$\mathrm{Substitue}\:{x}={u}+\mathrm{1}\:\mathrm{and}\:{y}={v}−\mathrm{4} \\ $$$$\frac{{dv}}{{du}}=\frac{{u}+{v}}{{u}−{v}} \\ $$$$ \\ $$

Answered by nburiburu last updated on 09/Jul/16

(x+y+3)dx+(−x+y+5)dy=0 { ((x+y+3=u+v)),((−x+y+5=u−v)) :} { ((u=y+4)),((v=x−1)) :} (u+v) dv+(u−v)du=0 homogeneus function u=t.v du=v.dt+t.dv v(t+1)dv+v(t−1)(vdt+tdv)=0 (t+1+t^2 −t)dv+(t−1)vdt=0 (1/v)dv =((1−t)/(1+t^2 )) dt ln v = atan(t)−(1/2)ln(1+t^2 )+c t=(u/v)=((y+4)/(x−1)) ; v=x−1 ln(x−1)=atan(((y+4)/(x−1)))−(1/2)ln((((x−1)^2 +(y+4)^2 )/((x−1)^2 )))+c and this should define implicitly y=f(x)

$$\left({x}+{y}+\mathrm{3}\right){dx}+\left(−{x}+{y}+\mathrm{5}\right){dy}=\mathrm{0} \\ $$$$\begin{cases}{{x}+{y}+\mathrm{3}={u}+{v}}\\{−{x}+{y}+\mathrm{5}={u}−{v}}\end{cases} \\ $$$$\begin{cases}{{u}={y}+\mathrm{4}}\\{{v}={x}−\mathrm{1}}\end{cases} \\ $$$$\left({u}+{v}\right)\:{dv}+\left({u}−{v}\right){du}=\mathrm{0} \\ $$$${homogeneus}\:{function} \\ $$$${u}={t}.{v} \\ $$$${du}={v}.{dt}+{t}.{dv} \\ $$$${v}\left({t}+\mathrm{1}\right){dv}+{v}\left({t}−\mathrm{1}\right)\left({vdt}+{tdv}\right)=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}+{t}^{\mathrm{2}} −{t}\right){dv}+\left({t}−\mathrm{1}\right){vdt}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{v}}{dv}\:=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$${ln}\:{v}\:=\:{atan}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$${t}=\frac{{u}}{{v}}=\frac{{y}+\mathrm{4}}{{x}−\mathrm{1}}\:;\:{v}={x}−\mathrm{1} \\ $$$${ln}\left({x}−\mathrm{1}\right)={atan}\left(\frac{{y}+\mathrm{4}}{{x}−\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{4}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\right)+{c} \\ $$$${and}\:{this}\:{should}\:{define}\:{implicitly}\:{y}={f}\left({x}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 09/Jul/16

Wow thanks why do we have tan^(−1) (t) i thought ((1 − t)/(1 + t^2 )) should be −(1/2)ln(1 + t^2 ) please am confuse in that integration ((1 − t)/(1 + t^2 ))dt why tan^(−1) (t) Thanks

$${Wow}\:{thanks} \\ $$$${why}\:{do}\:{we}\:{have}\:{tan}^{−\mathrm{1}} \:\left({t}\right) \\ $$$${i}\:{thought}\:\frac{\mathrm{1}\:−\:{t}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:{should}\:{be}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}\:+\:{t}^{\mathrm{2}} \right) \\ $$$${please}\:{am}\:{confuse}\:{in}\:{that}\:{integration} \\ $$$$\frac{\mathrm{1}\:−\:{t}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }{dt} \\ $$$${why}\:{tan}^{−\mathrm{1}} \left({t}\right)\:\:\:\:\:{Thanks} \\ $$

Commented by Tawakalitu. last updated on 09/Jul/16

Ohhhhh.... i got it now ((1 − t)/(1 + t^2 )) = (1/(1 + t^2 )) − (t/(1 + t^2 ))

$${Ohhhhh}....\:{i}\:{got}\:{it}\:{now} \\ $$$$ \\ $$$$\frac{\mathrm{1}\:−\:{t}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:−\:\frac{{t}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} } \\ $$

Commented by Tawakalitu. last updated on 09/Jul/16

please help with the latest question i asked. God bless

$${please}\:{help}\:{with}\:{the}\:{latest}\:{question}\:{i}\:{asked}. \\ $$$${God}\:{bless} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com