当f (t)是 二次多项式 时，解函数也是二次： \ [ { {y}_ {p}} (t)=a { {t}^ {2}}+bt+c\] 。 这些数字a，b，c是不确定的系数，可通过微分方程来确定它们。 对于任何恒定系数微分方程式（限定为特殊的f (t)），此方法均成功。 该方法可以更进一步，如果f (t)是多项式乘以指数，那么 \ [ { {y}_ {p}} (t)\] 具有相同的形式， \ [ { {y}_ {p}} (t)\] 中允许的t的最高幂次与f (t)中的相同。 只有发生共振，才允许在解中增加一个因子t。

2020年8月9日 · I can solve Euler $t^2 y''+t y' + y=0$ and $y''+p(t) y'(t) + q(t) y=0$ where it is possible to convert it to constant coefficient using known transformation. But this ODE is not one of these two types.

I wish to solve the following differential equation: $$ty' + 2y = t^2 - t + 1$$ for $t > 0$ and $y(1) = 2$. Seeing as I want to use the method of integrating factors, I divide everything by $...

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et −2et +et = 0. For y 2 we have y 2 0 = et(t + 1) and y 2 00 = et(t + 2). Substituting this into the given differential equation gives us et(t+2)−2et(t+1)+tet = 0. So y 1 and y 2 are both solutions to the differential equation. To see that they constitute a fundamental set of solutions, we examine their Wronskian: W(et,tet) = 1 y y 2 y 1 ...

2022年6月18日 · I have been trying to solve the differential equation $t y''(t) +2 y'(t) - ty(t)=0$ through the Frobenius method. WolframAlpha says the solution should be $ y(t) = c_1 \frac{e^{-t}}{t} + c_2\frac{e^t}{t}$. I got that $r=0 \lor r=-1$.

微分方程yy''-(y')^2=0的通解解法如下： 对一个微分方程而言，它的解会包括一些常数，对于n阶微分方程，它的含有n个独立常数的解称为该方程的通解。 例如：

2011年10月19日 · 2y-ty^2+e^t=5用隐函数求导法则：2y-ty^2+e^t=52y'-y^2-2ty*y'+e^t=0 (2-2ty)y'=y^2-e^ty'= (y^2-e^t)/ (2-2ty)

假设有 $n$ 个数据点 $(x_i, y_i), i=1,2,\cdots,n$，需要用最小二乘法拟合出一个曲线，可以使用 Matlab 中的 `polyfit` 函数实现。具体步骤如下： 1.

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