1) Find the solution to each equation with the following.Initial conditions:part a) dy. 11+ 4 = 1Y(2)=1part b) dy21:16Yt+2t4(-1)=1part c) yoxdy+ x = 5Y(5)=1

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1) Find the solution to each equation with the following. Initial conditions: part a) dy. 11+ y = 1 y(2) =1 part b) dy t²+1 = dt Yt+2t 4(-1)=1 part c) ydy + x = Y(5)=1

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.

Chapter11: Differential Equations

Section11.1: Solutions Of Elementary And Separable Differential Equations

Problem 19E: Find the particular solution for each initial value problem. dydx+3x2=2x;y(0)=5

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Question

1) Find the solution to each equation with the following.
Initial conditions:
part a) dy. 11+ 4 = 1
Y(2)=1
part b) dy
21:16
Yt+2t
4(-1)=1
part c) yox
dy
+ x = 5
Y(5)=1

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