If the solution y = y ( x ) of the differential equation ( x 4 + 2 x 3 + 3 x 2 + 2 x + 2 ) d y − ( 2 x 2 + 2 x + 3 ) d x = 0 satisfies y ( − 1 ) = − 4 π , then y ( 0 ) is equal to :

Question

If the solution y = y ( x ) of the differential equation ( x 4 + 2 x 3 + 3 x 2 + 2 x + 2 ) d y − ( 2 x 2 + 2 x + 3 ) d x = 0 satisfies y ( − 1 ) = − 4 π ​ , then y ( 0 ) is equal to :

TestHub · Accepted Answer

​ ∫ d y = ∫ x 4 + 2 x 3 + 3 x 2 + 2 x + 2 ( 2 x 2 + 2 x + 3 ) ​ d x y = ∫ ( x 2 + 1 ) ( x 2 + 2 x + 2 ) ( 2 x 2 + 2 x + 3 ) ​ d x y = ∫ x 2 + 2 x + 2 d x ​ + ∫ x 2 + 1 d x ​ y = tan − 1 ( x + 1 ) + tan − 1 x + C y ( − 1 ) = 4 − π ​ 4 − π ​ = 0 − 4 π ​ + C ⇒ C = 0 ⇒ y = tan − 1 ( x + 1 ) + tan − 1 x y ( 0 ) = tan − 1 1 = 4 π ​ ​

Mathematics - Differential Equation Question with Solution | TestHub

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