2. Find the explicit solution to the following initial value conditions: • (1 + x²)dy + x(1+4y²)dx = 0, y(1) = 0 dy dx dy = y² x²-1' y(2)=2 Xx +y=4x+1, y(1) = 8 dr ⚫ y' + tan xy = cos² x, y(0) = -1 (ex + y)dx+(2+x+ ye³)dy = 0, y(0) = 1 xy2 dy = y³-x³, y(1) = 2 dx ²-4-5y=0, y(1) = 0, y′(1) = 2 dt 4y" - y = xe/2, y(0) = 1, y'(0) = 0

2. Find the explicit solution to the following initial value conditions: • (1 + x²)dy + x(1+4y²)dx = 0, y(1) = 0 dy dx dy = y² x²-1' y(2)=2 Xx +y=4x+1, y(1) = 8 dr ⚫ y' + tan xy = cos² x, y(0) = -1 (ex + y)dx+(2+x+ ye³)dy = 0, y(0) = 1 xy2 dy = y³-x³, y(1) = 2 dx ²-4-5y=0, y(1) = 0, y′(1) = 2 dt 4y" - y = xe/2, y(0) = 1, y'(0) = 0

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter6: Applications Of The Derivative

Section6.CR: Chapter 6 Review

Problem 36CR

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