Question

Y''*x*ln(x)-y'=0 диф уравнения

Answer 1

Ответ:

Сделаем замену z=y', w=lnx

z'xlnx=z

dz/z=dx/xlnx

dz/z=dw/w

lnz=lnw+lnC, C=const

z=wC

y'=Clnx

y=Cxlnx-Cx+D, D=const

y'=Clnx

y=Cxlnx-

Пошаговое объяснение:

Answer 2

The given differential equation is a second-order linear homogeneous ordinary differential equation. It's written as:

$y'' \cdot x \cdot \ln(x) - y' = 0$.

Let's solve this equation step by step:

Step 1: Rewrite the equation in standard form. Divide both sides of the equation by $x \cdot \ln(x)$: $\frac{y''}{y'} = \frac{1}{x \cdot \ln(x)}$.

Step 2: Integrate both sides with respect to $x$. $\int \frac{y''}{y'} \, dx = \int \frac{1}{x \cdot \ln(x)} \, dx$.

The left side can be integrated using the substitution $u = y'$, so $du = y'' \, dx$: $\int \frac{du}{u} = \int \frac{1}{x \cdot \ln(x)} \, dx$.

This simplifies to: $\ln|u| = \int \frac{1}{x \cdot \ln(x)} \, dx$.

Step 3: Solve the integral on the right-hand side. The integral on the right side is not elementary, meaning it doesn't have a simple closed-form solution in terms of basic functions. You might need to use more advanced techniques like integration by parts or consider it in terms of special functions.

Step 4: Solve for $u$ (which is $y'$). Exponentiate both sides of the equation: $|u| = e^{\int \frac{1}{x \cdot \ln(x)} \, dx}$.

Since $u$ represents the derivative of $y$, we can rewrite this as: $y' = C \cdot e^{\int \frac{1}{x \cdot \ln(x)} \, dx}$.

Where $C$ is an arbitrary constant.

Step 5: Integrate again to find $y$. Integrate $y'$ with respect to $x$ to find $y$: $y(x) = \int C \cdot e^{\int \frac{1}{x \cdot \ln(x)} \, dx} \, dx + D$.

Here, $D$ is another arbitrary constant.

The integral in the expression for $y(x)$ might not have a simple closed-form solution, so in practice, you might leave the solution in terms of the integrals.

Please note that solving this kind of equation might involve complex mathematical techniques, and the resulting solution could be quite involved due to the nature of the integral. If you have specific initial or boundary conditions, you can use them to determine the constants $C$ and $D$ and obtain a more concrete solution.