MAPLE求解ODE初值问题疑问：矛盾情况与通解可能性

Hey there! Let's break down this problem step by step— that confusing 4=0 contradiction you're hitting almost certainly comes from a small misstep in either simplifying the equation or entering it into Maple. Let's start with the basics.

Step 1: Correctly Rewrite the ODE in Standard Linear Form #

Your original equation is:
$$(t+1)\frac{dy}{dt}-2(t^{2+t)y=\frac{e}{t^2}}{t+1}$$
Since $t > -1$, we can safely divide every term by $(t+1)$ to get the standard linear ODE form $\frac{dy}{dt} + P(t)y = Q(t)$:
$$\frac{dy}{dt} - 2t y = \frac{e^{t2}}{(t+1)^2}$$
Note: This is where I suspect you might have made a mistake— if you forgot to divide the right-hand side by $(t+1)$ a second time, your standard form would be wrong, leading to unexpected results when plugging in the initial condition.

Step 2: Manually Derive the General Solution #

For linear ODEs, we use an integrating factor $\mu(t) = \exp\left(\int P(t) dt\right)$. Here, $P(t) = -2t$, so:
$$\mu(t) = \exp\left(\int -2t dt\right) = e^{-t2}$$
Multiply the entire standard form by $\mu(t)$:
$$e^{-t2}\frac{dy}{dt} - 2t e^{-t2} y = \frac{1}{(t+1)^2}$$
The left-hand side is the derivative of $\mu(t)y(t)$, so we integrate both sides:
$$\int \frac{d}{dt}\left(e^{-t2}y(t)\right) dt = \int \frac{1}{(t+1)^2} dt$$
Calculating the integrals gives:
$$e^{-t2}y(t) = -\frac{1}{t+1} + C$$
Solve for $y(t)$ to get the general solution:
$$y(t) = e^{t2}\left(C - \frac{1}{t+1}\right)$$

Step 3: Apply the Initial Condition $y(0)=5$ #

Plug $t=0$ and $y=5$ into the general solution:
$$5 = e^{0}\left(C - \frac{1}{0+1}\right)$$
$$5 = 1 \cdot (C - 1)$$
$$C = 6$$
So the unique solution to your initial value problem is:
$$y(t) = e^{t2}\left(6 - \frac{1}{t+1}\right)$$

Step 4: Correct Maple Implementation #

To avoid errors in Maple, input the original ODE directly (no need to pre-simplify it!) along with the initial condition:

# Define the ODE
ode := (t+1)*diff(y(t), t) - 2*(t^2 + t)*y(t) = exp(t^2)/(t+1);

# Define the initial value problem
ivp := {ode, y(0) = 5};

# Solve the IVP
dsolve(ivp);

Running this code should return the exact solution we derived manually:
$$y(t) = e^{t2}\left(6 - \frac{1}{t+1}\right)$$

Why Did You Get a 4=0 Contradiction? #

Most likely, you made an error when simplifying the ODE to standard form (e.g., miscalculating the right-hand side) or when entering the equation into Maple. For example, if you accidentally wrote the right-hand side as $\frac{e^{t2}}{t+1}$ instead of $\frac{e^{t2}}{(t+1)^2}$ in your standard form, your general solution would be incorrect, leading to a contradiction when plugging in $y(0)=5$.

Key Takeaway #

Your ODE absolutely has a general solution, and the initial value problem has a unique solution (since $P(t)$ and $Q(t)$ are continuous for $t > -1$, satisfying the existence-uniqueness theorem for linear ODEs). The contradiction was just a sign of a small algebraic or input error.

内容的提问来源于stack exchange，提问作者Bazfred