Question

Y'+y/2x =x^2, y(1)=1

Answer 1

Ответ:

$\displaystyle y(x)= \frac27x^3+ \frac5{7\sqrt x}$

Пошаговое объяснение:

Найдем сначала общее решение дифференциального уравнения

$\displaystyle y'(x)+\dfrac{y(x)}{2x}=x^2\\\frac{dy(x)}{dx}+\frac{y(x)}{2x}=x^2\\\sqrt x\frac{dy(x)}{dx}+\frac{y(x)}{2\sqrt x}=x^{\frac52}\\\sqrt x\frac{dy(x)}{dx}+\frac{d2\sqrt x}{dx}y(x)=x^{\frac52}\\\frac d{dx}(\sqrt x y(x))=x^\frac{5}{2}\\\frac d{dx}(\sqrt x y(x))dx=x^\frac{5}{2}dx\\\int\frac d{dx}(\sqrt x y(x))dx=\int x^\frac{5}{2}dx\\\sqrt xy(x)=\frac27x^\frac72+c\\y(x)=\frac27x^3+\frac c{\sqrt x}$

Чтобы найти частное решение, надо решить уравнение

$\displaystyle y(1)=1\\y(1)=\frac271^3+\frac c{\sqrt 1}=1\\\frac271^3+\frac c{\sqrt 1}=1\\\frac27+c=1\\c=1-\frac27\\c=\frac57$

Подставим вместо с это значение и получим решение

Answer 2

This is a first-order linear ordinary differential equation, and we can solve it using an integrating factor. The equation is in the form:

$\frac{dy}{dx} + \frac{y}{2x} = x^2$

To solve this equation, we'll follow these steps:

Step 1: Write the equation in standard form $\frac{dy}{dx} = -\frac{y}{2x} + x^2$

Step 2: Identify the integrating factor The integrating factor ($IF$) for this equation is given by the exponential of the integral of the coefficient of $y$ with respect to $x$: $IF = e^{\int -\frac{1}{2x} \,dx} = e^{-\frac{1}{2} \ln(x)} = \frac{1}{\sqrt{x}}$

Step 3: Multiply both sides of the equation by the integrating factor $\frac{1}{\sqrt{x}} \cdot \frac{dy}{dx} + \frac{y}{2x\sqrt{x}} = x^{\frac{3}{2}}$

Step 4: Recognize the left side as the derivative of $\left(\frac{y}{\sqrt{x}}\right)$ $\frac{d}{dx} \left(\frac{y}{\sqrt{x}}\right) = x^{\frac{3}{2}}$

Step 5: Integrate both sides $\int \frac{d}{dx} \left(\frac{y}{\sqrt{x}}\right) \,dx = \int x^{\frac{3}{2}} \,dx$ $\frac{y}{\sqrt{x}} = \frac{2}{5} x^{\frac{5}{2}} + C$

Step 6: Solve for $y$ $y = \frac{2}{5} x^3\sqrt{x} + C\sqrt{x}$

Step 7: Use the initial condition to solve for $C$ Given the initial condition $y(1) = 1$, we can substitute it into the equation: $1 = \frac{2}{5} \cdot 1^3\sqrt{1} + C\sqrt{1}$ $1 = \frac{2}{5} + C$ $C = \frac{3}{5}$

So, the solution to the initial value problem is: $y = \frac{2}{5} x^{\frac{7}{2}} + \frac{3}{5} x^{\frac{1}{2}}$