Solve the initial value problem 8(t+1) dy/dt
−6y=12t for t>−1 with y(0)=17. Find the integrating factor, u(t)=
EVESUNER8570
To solve this initial value problem, we first need to put the equation in the standard form of a linear first-order ordinary differential equation: dy/dt + P(t)y = Q(t). The given equation can be rearranged to this form by dividing both sides by 8(t+1):
dy/dt - (6/8(t+1))y = (12t)/(8(t+1))
Comparing this to the standard form, we have P(t) = -(6/8(t+1)) and Q(t) = (12t)/(8(t+1)).
To find the integrating factor, we use the formula u(t) = e^(∫P(t)dt). In this case, P(t) = -(6/8(t+1)), so we have:
∫P(t)dt = ∫-(6/8(t+1))dt
= -(6/8)∫(1/(t+1))dt
= -(6/8)ln|t+1| + C,
where C is the constant of integration.
Therefore, the integrating factor is u(t) = e^(-(6/8)ln|t+1| + C)
= e^(-(3/4)ln|t+1| + C)
= e^(ln|t+1|^(-3/4) + C)
= e^(ln|t+1|^(-3/4)) * e^(C)
= |t+1|^(-3/4) * e^(C).
Since ln|t+1| and C are constant terms, e^(ln|t+1|^(-3/4)) * e^(C) is also a constant.
Therefore, the integrating factor for this initial value problem is u(t) = |t+1|^(-3/4) * constant.
To find the integrating factor, we look at the coefficient of dy/dt in the given differential equation, which is 8(t + 1). We can see that this coefficient is a function of t.
The integrating factor, u(t), for an equation of the form dy/dt + P(t)y = Q(t), is given by u(t) = e^(integral of P(t) dt).
In this case, we have P(t) = 8(t + 1). So, we calculate the integral of P(t) dt:
∫P(t) dt = ∫8(t + 1) dt
= 8∫(t + 1) dt
= 8(∫t dt + ∫1 dt)
= 8(1/2 t^2 + t) + C
= 4 t^2 + 8t + C
Therefore, the integrating factor, u(t), is given by u(t) = e^(4t^2 + 8t + C), where C is a constant.