微分方程$(4+t^2)y'(t)+2ty(t)=4t$求解方法咨询

Hey Nora, no stress—let's walk through solving this differential equation together. It's a linear first-order ordinary differential equation, so the integrating factor method is perfect here. Let's break it down step by step:

1. Rewrite the equation into standard linear form
   First, we need to get it into the form $y'(t) + P(t)y(t) = Q(t)$. Divide every term by $(4+t^2)$:
   $$y'(t) + \frac{2t}{4+t^2}y(t) = \frac{4t}{4+t^2}$$

2. Calculate the integrating factor
   The integrating factor $\mu(t)$ is given by $\mu(t) = e^{\int P(t)dt}$. Here, $P(t) = \frac{2t}{4+t^2}$. Let's compute the integral:

   • Let $u = 4 + t^2$, so $du = 2t dt$. The integral becomes $\int \frac{du}{u} = \ln(u) + C = \ln(4+t^2)$ (since $4+t^2$ is always positive, we can drop the absolute value).
   • So $\mu(t) = e^{\ln(4+t2)} = 4 + t^2$.

3. Multiply through by the integrating factor
   Multiply every term in the standard form by $\mu(t)$:
   $$(4+t^2)y'(t) + 2ty(t) = 4t$$
   Wait a minute—this is exactly your original equation! The left-hand side is the derivative of the product $\mu(t)y(t)$, so we can rewrite this as:
   $$\frac{d}{dt}\left[(4+t^2)y(t)\right] = 4t$$

4. Integrate both sides
   Integrate the left side (which just gives us the product we had) and the right side:
   $$(4+t^2)y(t) = \int 4t dt$$
   The integral of $4t$ is $2t^2 + C$, where $C$ is the constant of integration.

5. Solve for $y(t)$
   Divide both sides by $(4+t^2)$ to isolate $y(t)$:
   $$y(t) = \frac{2t^2 + C}{4+t^2}$$
   You can also simplify this if you want: split the fraction to get $y(t) = 2 - \frac{8 - C}{4+t^2}$ (since $\frac{2t^2 + C}{4+t^2} = \frac{2(t^2+4) + (C-8)}{t^2+4} = 2 + \frac{C-8}{t^2+4}$). Either form is correct.

6. Verify the solution (optional but helpful)
   To make sure this is right, let's plug $y(t)$ back into the original equation. Compute $y'(t)$ using the quotient rule:
   $$y'(t) = \frac{4t(4+t^2) - (2t^2 + C)(2t)}{(4+t²⁾2} = \frac{16t + 4t^3 - 4t^3 - 2Ct}{(4+t²⁾2} = \frac{16t - 2Ct}{(4+t²⁾2}$$
   Now substitute into the left-hand side of the original equation:
   $$(4+t^2)y'(t) + 2ty(t) = \frac{16t - 2Ct}{4+t^2} + \frac{2t(2t^2 + C)}{4+t^2} = \frac{16t - 2Ct + 4t^3 + 2Ct}{4+t^2} = \frac{4t^3 + 16t}{4+t^2} = 4t$$
   That's exactly the right-hand side of the original equation, so our solution checks out!

If you got stuck somewhere along the way, feel free to ask more questions about the steps that tripped you up.

备注：内容来源于stack exchange，提问作者Nora