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Quiz-6 / Re: Q6 TUT 0101
« on: November 18, 2018, 12:57:01 PM »
My answer is different.
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Quiz-6 / Re: Q6 TUT 0401
« on: November 18, 2018, 12:50:58 PM »
My answer to c is slightly different.
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Quiz-4 / Re: Q4 TUT 0601
« on: October 26, 2018, 09:49:14 PM »
For the homogeneous equation,
r² + 1 = 0
r = i or −i
So, the complementary solution is y = C1cost + C2sint
W[y1,y2] =y1y2' − y1'y2 = cos²t + sin²t = 1
u1(x)=−∫(y2(x)g(x)/W)dx
= −∫(sint tant) dt
= sint − ln(tant+sect)
u2(x)=∫(y1(x)g(x)/W)dx
= ∫(cost tant)dt
= ∫sint dt
= −cos t
So, a particualr solution is
y = u1y1+u2y2
=[sint - ln(tant+sect)]cost − costsint
=−ln(tant+sect)cost
So,the general solution is
y = C1cost + C2sint −ln(tant+sect)cost
r² + 1 = 0
r = i or −i
So, the complementary solution is y = C1cost + C2sint
W[y1,y2] =y1y2' − y1'y2 = cos²t + sin²t = 1
u1(x)=−∫(y2(x)g(x)/W)dx
= −∫(sint tant) dt
= sint − ln(tant+sect)
u2(x)=∫(y1(x)g(x)/W)dx
= ∫(cost tant)dt
= ∫sint dt
= −cos t
So, a particualr solution is
y = u1y1+u2y2
=[sint - ln(tant+sect)]cost − costsint
=−ln(tant+sect)cost
So,the general solution is
y = C1cost + C2sint −ln(tant+sect)cost
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Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 10:27:37 PM »
By the way, I think my answer to the original question is correct.
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Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 09:18:48 PM »
Happy Thanksgiving.
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