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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 217 times)

magmichele12

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on: May 27, 2021
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) + cos(x)
◦ y = + cos3 x - sin(x) + cos(x)
◦ y = + cos3 x - sin(x) + 7 cos(x)
◦ y = -cos3 x - sin(x) + 7 cos(x)
◦ y = cos3 x - sin(x) + cos(x)
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Marked as best answer by magmichele12 on May 27, 2021

zacnyjessica

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Reply #1 on: May 27, 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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magmichele12

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Reply 2 on: May 27, 2021
:D TYSM

dawsa925

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

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