Question Number 102515 by bemath last updated on 09/Jul/20

((x/y))y′= ((2y^2 +1)/(x+1))

Answered by bobhans last updated on 09/Jul/20

(dy/(y(2y^2 +1))) = (dx/(x(x+1)))  (1) (1/(y(2y^2 +1))) = (A/y) + ((By+C)/(2y^2 +1))  1 = A(2y^2 +1)+(By+C)y   ⇒ { ((A=1)),((1=3.1−(−B+C))),((1=3.1+B+C)) :}  −2 = B−C ∧−2=B+C  ⇒B=−2 ; C=0  ∫(dy/y)−∫((2y dy )/(2y^2 +1)) = ∫ (dx/x) −∫ (dx/(x+1))  ln(y)−(1/2)ln(2y^2 +1) = ln∣C((x/(x+1)))∣   ln∣(y/(√(2y^2 +1)))∣ = ln ∣((Cx)/(x+1)) ∣   (y/(√(2y^2 +1))) = ((Cx)/(x+1)) ⊕

Answered by petrochengula last updated on 09/Jul/20

(dy/(y(2y^2 +1)))=(dx/(x(x+1)))  integrate both sides  ∫(dy/(y^3 (2+(1/y^2 ))))=∫(dx/(x^2 (1+(1/x))))  −(1/2)ln(((2y^2 +1)/y^2 ))=ln∣((Cx)/(1+x))∣+  ln(√(y^2 /(2y^2 +1)))=ln∣((Cx)/(1+x))∣  (y/(√(2y^2 +1)))=((Cx)/(1+x))