Find the solution to this equation with 2 given solutions help ps?

Question

I am stuck help me please,

the equation is y'+6yt=12t
initial conditions are y(0)=4 & y(0)=0

I have solved for my equation and have come to:

y e^6t = 12 ( 1/6 e^6t - 1/36 e^6t) + C

But i can't seem to find anything that works from here, have i solved my first order differential wrong?

I could reduce the equation on the RHS and then divide both sides by e^6t which would leave me with 2 - 1/3 + C/e^6t but i think thats wrong.


THank you in advance

Answer 1

y' + 6y t = 12t ----------(1)



Linear equation of first order of the form y' + p(t) y = q(t)

P(t) = 6t

q(t) = 12t



The integrating factor is e^∫ p(t) dt = e^∫ 6t dt = e^(3t^2)

Multiply equation (1) by e^(3t^2)



e^(3t^2) y' + 6y t e^(3t^2) = 12 t e^(3t) --------(2)

The left hand side is d/dt ( y e^(3t^2) ) or (y e^(3t^2) )'



(y e^3t^2)' = 12 t e^(3t^2)

Integrate both sides



y e^(3t^2) = ∫ 12 t e^(3t^2) dt



let u = 3t^2

du = 6 t dt

t dt = (1/6) du

12 ∫ t e^(3t^2) dt = (12)(1/6) ∫ e^u du = 2 e^u = 2 e^(3t^2)



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

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

y(0) = 4

4 = 2 + C

C = 2



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



y(0) = 0

0 = 2 + C /1

C = 2

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

Answer 2

dy/dt = 6t(2 - y) =>

dy/(2-y) = 6t dt =>

ln|2-y| = 3t^2 + C1 =>

2 - y = ce^(3t^2).

y(0) = 4 => c = -2.

If y(0) were instead 0, you'd have c = 2.

You can't have y(0) = 4 and y(0) = 0 at the same time!

Answer 3

Assuming you did not mean dy/dt + 6y(t) on the left, but actually meant

dy/dt + 6t*y(t) = 12t

dy/dt = -6t(y – 2) ….separable

dy/(y – 2) = -6t dt

ln(y – 2) = -3t^2 + k

y – 2 = e^k*e^(-3t^2) = C e^(-3t^2)



Only one initial condition required; Lets take y(0)=4

4 = 2 + C

y = 2[e^(-3t^2) + 1]

Answer 4

　

For a 1st degree polynomial, you need just 1 initial condition.

Besides, how can y(0) be both = 0 and 4?



y' + 6ty = 12t



Integrating factor: e^∫ (6t) dt = e^(3t²)



e^(3t²)y' + 6te^(3t²)y = 12te^(3t²)

d/dt (e^(3t²)y) = 12te^(3t²)

d (e^(3t²)y) = 12te^(3t²) dt

∫ d (e^(3t²)y) = ∫ 12te^(3t²) dt

e^(3t²)y = 2e^(3t) + C

y = 2 + Ce^(−3t²)



y(0) = 2 + Ce^0 = 2 + C



For y(0) = 0 ----> C = −2 ----> y = 2 − 2e^(−3t²)

For y(0) = 4 ----> C = 2 -----> y = 2 + 2e^(−3t²)

Answer 5

Grammatically challenged professional Internet troll is also mathematically challenged:



"Find the solution to this equation with 2 given solutions help ps? I am stuck help me please, the equation is y'+6yt=12t initial conditions are y(0)=4 & y(0)=0 I have solved for my equation and have come to: y e^6t = 12 ( 1/6 e^6t - 1/36 e^6t) + C But i can't seem to find anything that works from here, have i solved my first order differential wrong? I could reduce the equation on the RHS and then divide both sides by e^6t which would leave me with 2 - 1/3 + C/e^6t but i think thats wrong. THank you in advance"

• https://answersrip.com/question/index?qid=20170416074813AAX8iMC



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