Question Number 104342 by bemath last updated on 21/Jul/20

solve x (d^2 y/dx^2 )−(dy/dx)−4x^3 y = 8x^3 sin(x^2 )

Answered by bramlex last updated on 21/Jul/20

multiplying both sides by x  ⇒x^2 .(d^2 y/dx^2 )−x(dy/dx)−4x^4 y=8x^4 sin (x^2 )...(1)  set x^2  = z → (dz/dx) = 2x   (dy/dx) = (dy/dz).(dz/dx) = 2x (dy/dz)→ (dy/dx) = 2x.(dy/dz)...(2)  (d^2 y/dx^2 ) = (d/dx)(2x (dy/dz)) = 2(dy/dz)+2x (d/dx)((dy/dx))           = 2(dy/dz) +2x (d/dz)((dy/dz)).(dz/dx)...(3)  using (2) &(3) in (1)  (→)x^2 [2(dy/dx)+4x^2  (d^2 y/dz^2 ) ]−x(2x(dy/dz))  −4x^4 y = 8x^4  sin (x^2 )  (→) 4x^4  ((d^2 y/dz^2 ) −y) = 8x^4  sin (z)  (→) (d^2 y/dz^2 ) −y = 2sin (z)   (→) (D^2 −1)y = 2sin (z)  Homogenous part  y_c  = C_1 e^(−x) +C_2 e^x   particular integral  y_p = (1/(D^2 −1))(2sin (z))= −sin (z)=−sin (x^2 )  General solution  ∴ y = C_1 e^(−x) +C_2 e^x −sin (x^2 ) ★

Commented bybemath last updated on 21/Jul/20

great...nice