integrating factor problem with given solution?
the function y(t)=t^2 is a solution of y′−[2/t]y=0. Using this information, what is the general solution of the differential equation? If you need an arbitrary constant in your answer, use a lower-case "c".
i understand how to do a normal integrating factor problem, but i don't get where the function given (t^2) comes in
the answer however, is ct^2, not t^2+c
- Engr. RonaldLv 78 years agoBest Answer
let P(x) = - 2/t, Q(x) = 0
IF = e^[-2∫dt/t] = e^[-2ln(t)] = e^[ln(t)^(-2)] = t^(-2)
y/t^2 =∫ t^2(0)dt
y/t^2 = C
y = Ct^2 answer//
- bobbyjimthefirstLv 58 years ago
The given answer is just one solution of the Bernoulli differential equation, if you can do the integrating factor method, you can find the general solution that way.
However, the information y(t) = t² satisfies the equation is very helpful.
What you can do is consider the way this differentiates:
d[t²] = 2t dt.
This tells you that, if you were to multiply through by some constant, say, c, it would remain after differentiation, and then cancel itself out in the differential equation.
d[ct²] = 2ct dt.
d/dt[ct²] = 2ct.
2ct - [2 / t] ct² = 0.
Factoring c and simplifying:
c (2t - 2t) = 0.
The bracket is clearly zero, hence, the value of c as a multiplier is irrelevant.
You could not do this with y(t) = t² + c, because, on differentiation, the constant disappears:
d[t² + c] = 2t dt.
So you cannot factor it out, and it will not be irrelevant.
In fact, you will get:
2t - [2 / t] (t² + c) = 2t - 2t + 2c/t = 2c / t
Clearly, 2c / t ≠ 0 for every constant c.
You can check the answer by using the integrating factor method.
This is a Bernoulli differential equation, with parameters:
n = 0.
P(t) = -2 / t.
Q(t) = 0.
The integrating factor is IF = e^(∫ P(t) dt).
First, do the integral:
∫ P(t) dt = ∫ -2dt / t = -2 ∫ dt / t = -2 ln(t) = ln(1 / t²).
Then, calculate the integrating factor:
IF = e^(∫ P(t) dt) = e^(ln(1 / t²)) = 1 / t².
Multiply the DE by the IF:
(dy/dt) / t² - [2 / t] [y / t²] = 0 / t².
(dy/dt) / t² - 2y / t³ = 0.
The LHS is now a complete product derivative of y * IF:
d/dt [y / t²] = 0.
Apply integrals with respect to t:
∫ d/dt [y / t²] dt = ∫ 0 dt.
Cancel derivative with integral:
y / t² = ∫ 0 dt.
The integral of 0 is just a constant, say, c:
y / t² = c.
Multiply by t²:
y = ct².
- davidLv 68 years ago
using the integrating factor t^(-2)
y'/t^2 - (2y/t^3) = 0
now we can integrate
y/t^2=c, for some constant c
y=ct^2 is the general solution
- cidyahLv 78 years ago
If y=t^2 is a solution of y'-(2/t) y=0, it should satisfy the differential equation.
y' -(2/t)y = 2t -(2/t) t^2 = 2t-2t=0
The solution therefore is y=t^2+c
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- billyLv 43 years ago
your integrating factor is actual. yet we could think of related to the final form first: y' + a(x)y=b(x) to that end a(x)=2x and b(x)=0 appropriate? we could call the integrating factor M(x), which to that end you solved wisely to be M(x)=e^(x^2). then what you get is that: y=( crucial( b(x) M(x) dx ) + C)/M(x) yet considering the fact that b(x)=0 and you have M(x) above: y= C/M(x) use you preliminary situations y(0)=2 (which will desire to be C=-2) y=-2 e^(x^2) = -2e^(-x^2)