Question

(x^2 + 1) dy/dx + 3x^3y = 6xe^(-3x^2/2); y(0) = 1

          (x^2 + 1) dy/dx + 3x^3y = 6xe^(-3x^2/2); y(0) = 1

(x^2 + 1) dy/dx + 3x^3y = 6xe^(-3x^2/2); y(0) = 1

Added by Alan G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

01/13/2023

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dy = 3x2 + 3x3 = 6xe Show more…

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(x^2 + 1) dy/dx + 3x^3y = 6xe^(-3x^2/2); y(0) = 1

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