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m_a_m_a [10]
4 years ago
15

Solve the differential equation t dy/dt + dy/dt = te ^y

Mathematics
1 answer:
LuckyWell [14K]4 years ago
5 0

The answer is

t dy / dt + dy / dt = te ^ y

I apply common factor                1/dt*(tdy+dy)=te^y

I pass "dt"                    tdy+dy=(te^y)*dt

I apply common factor                (t+1)*dy=(te^y)*dt

I pass "e^y"                         (1/e^y)dy=((t+1)*t)*dt

I apply integrals                ∫ (1/e^y)dy= ∫ (t^2+t)*dt

by property of integrals  ∫ (1/e^y)dy= ∫ (e^-y)dy

∫ (e^-y)dy= ∫ (t^2+t)*dt

I apply integrals

-e^-y=(t^3/3)+(t^2/2)+C

I apply natural logarithm to eliminate "e"

-ln (e^-y)=-ln(t^3/3)+(t^2/2)+C

y=ln(t^3/3)+(t^2/2)+C

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Answer:

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Step-by-step explanation:

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P(S=2) = 2/35

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P(S=6)/3 = (1/3)(6/35) = 2/35

Because for S=6: (2,4) (3,3) (4,2)

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P(getting doubles = 2/35 + 4/105 + 2/35 + 8/35 = 40/105 = 8/21

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