Let y = y(x) be the solution of the differential equation (x² + 1)y′ − 2xy = (x⁴ + 2x² + 1)cosx, y(0) = 1. Then ³∫₋₃ y(x) dx is:
Question Let y = y ( x ) y = y(x) be the solution of the differential equation ( x 2 + 1 ) y ′ − 2 x y = ( x 4 + 2 x 2 + 1 ) cos x , with y ( 0 ) = 1 y(0) = 1 . Then evaluate: ∫ − 3 3 y ( x ) d x Question Image Short Solution Rewrite the ODE in standard linear form: y ′ − 2 x x 2 + 1 y = ( x 4 + 2 x 2 + 1 ) cos x x 2 + 1 Find the integrating factor : μ ( x ) = e ∫ − 2 x x 2 + 1 d x = e − ln ( x 2 + 1 ) = 1 x 2 + 1 Multiply the equation by μ ( x ) \mu(x) : d d x ( y x 2 + 1 ) = cos x Integrate: y x 2 + 1 = sin x + C Apply y ( 0 ) = 1 y(0)=1 : 1 1 = 0 + C ⟹ C = 1 Hence: y ( x ) = ( x 2 + 1 ) ( sin x + 1 ) Compute the definite integral: I = ∫ − 3 3 y ( x ) d x = ∫ − 3 3 ( x 2 + 1 ) ( sin x + 1 ) d x Image Solution Conclusion Break the integral: I = ∫ − 3 3 ( x 2 + 1 ) sin x d x + ∫ − 3 3 ( x 2 + 1 ) d x First term: ( x 2 + 1 ) sin x (x^{2}+1)\sin x is an odd function (because x 2 + 1 x^{2}+1 is eve...