Solving ty' - y = t^3 - 2t

[tandoorichicken]

I'm having a little trouble with this differential equation. Usually I have no problem separating out the y's and the t's but this one is different:

ty' - y = t^3 - 2t on (0, $\infty$)

A little hep please?

 

[dextercioby]

Since $t\neq 0$,u can, perform a division and obtain

$$\frac{dy}{dt}-\frac{y}{t}=t^{2}-2$$

Solve this nonhomogenous linear first order ODE,knowing that the homogenous equation is separable...


Daniel.

 

[thepatientmental]

Sure, I would be happy to provide some assistance with this differential equation. First, let's rewrite the equation in a more standard form:

y' - (1/t)y = t^2 - 2

To solve this equation, we can use the method of integrating factors. The integrating factor is defined as e^(∫(1/t)dt), which in this case is e^(ln|t|) = t. Multiplying both sides of the equation by t, we get:

ty' - y = t^3 - 2t

ty' - (ty) = t^4 - 2t^2

Applying the product rule on the left side, we get:

t(y' - y) = t^4 - 2t^2

Integrating both sides with respect to t, we get:

∫t(y' - y)dt = ∫(t^4 - 2t^2)dt

Using integration by parts on the left side, we get:

ty - ∫ydt = ∫(t^4 - 2t^2)dt

Simplifying, we get:

ty - y = t^5/5 - 2t^3/3 + C

Finally, solving for y, we get the general solution:

y = (t^5/5 - 2t^3/3 + C)/t

To find the particular solution, we need to use the initial condition of y(0) = 0. Plugging in t = 0 and y = 0 into the general solution, we get C = 0. Therefore, the particular solution to this differential equation is:

y = (t^5/5 - 2t^3/3)/t = t^4/5 - 2t^2/3

I hope this helps! Let me know if you have any further questions.