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Author Question: Solve the following initial-value problem:cos(x)y' + sin(x)y = 2 cos3(x) sin(x) - 1, y = 3, 0 x . (Read 125 times)

jace

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on: May 26, 2021

Question 1

Solve the following  initial-value problem   = 9.8 - 0.196v,  v(0) = 48.
◦ v = 50 + 2e-0.196t
◦ v = 50 - e-0.196t
◦ v = 50 - 2e-0.196t
◦ v = 100 - 4e-0.196t
◦ v = 100 + 2e-0.196t

Question 2

Solve the following initial-value problem:
cos(x)y' + sin(x)y = 2 cos3(x) sin(x) - 1,  y = 3,  0 ≤  x ≤ .
◦ y = - cos3 x - sin(x) + 7 cos(x)
◦ y = cos3 x - sin(x) + cos(x)
◦ y = cos3 x - sin(x) + 7 cos(x)
◦ y = cos3 x + sin(x) + cos(x)
◦ y = - cos3 x - sin(x) + cos(x)
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Marked as best answer by jace on May 26, 2021

cegalasso

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Reply #1 on: May 26, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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jace

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Reply 2 on: May 26, 2021
:D TYSM

Liddy

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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