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 119002 by benjo_mathlover last updated on 21/Oct/20

(dy/dx) = ((xy)/(x^2 +1)) and y((√(15))) = 2

$$\frac{{dy}}{{dx}}\:=\:\frac{{xy}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{y}\left(\sqrt{\mathrm{15}}\right)\:=\:\mathrm{2} \\ $$

Answered by bramlexs22 last updated on 21/Oct/20

 (dy/y) = ((x dx)/(x^2 +1)) ⇒ ∫ (dy/y) = ∫ ((x dx)/(x^2 +1))  ln (y) = (1/2)ln (x^2 +1)+C   ln (y)= ln (λ(√(x^2 +1))) ⇒y = λ(√(x^2 +1))  ⇒2 = λ(√(16)) ; λ=(1/2)  therefore y = ((√(x^2 +1))/2)

$$\:\frac{{dy}}{{y}}\:=\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:\int\:\frac{{dy}}{{y}}\:=\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{ln}\:\left({y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{C}\: \\ $$$$\mathrm{ln}\:\left({y}\right)=\:\mathrm{ln}\:\left(\lambda\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:\Rightarrow{y}\:=\:\lambda\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\:=\:\lambda\sqrt{\mathrm{16}}\:;\:\lambda=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${therefore}\:{y}\:=\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com