Question

Solve the differential equation xdy/dx+y=x3

Answer 1

Answer on Question #69634 – Math – Differential Equations

Question

Solve the differential equation

$$x dy/dx + y = x^3$$

Solution

Divide both sides of the linear differential equation

$$xy' + y = x^3$$

by $x$ and obtain

$$y' + \frac{y}{x} = x^2$$

Let

$$y = UV.$$

Then

$$\begin{aligned} y' &= U'V + V'U \\ U'V + V'U + \frac{UV}{x} &= x^2 \\ U'V + U\left(V' + \frac{V}{x}\right) &= x^2. \end{aligned}$$

If

$$V' + \frac{V}{x} = 0,$$

then

$$\begin{aligned} U'V + U \cdot 0 &= x^2, \\ U'V &= x^2. \end{aligned}$$

Solving the equation (2)

$$\begin{aligned} V' + \frac{V}{x} &= 0, \\ \frac{dV}{dx} &= -\frac{V}{x'}, \\ \int \frac{dV}{V} &= -\int \frac{dx}{x'}, \\ \ln V &= \ln \frac{1}{x} + \ln C. \end{aligned}$$

Let $\ln C = 0$, then

$$V = \frac{1}{x}.$$

Substituting (4) into equation (3)

$$U'V = x^2$$

one gets

$$U _ {x} ^ {\prime} = x ^ {2},$$ $$U ^ {\prime} = x ^ {3},$$ $$U = \int x ^ {3} d x,$$ $$U = \frac {x ^ {4}}{4} + C,$$

where $C$ is an integration constant.

Substituting (4) and (5) into (1) one gets the solution of the initial differential equation:

$$y = U V \Rightarrow y = \left(\frac {x ^ {4}}{4} + C\right) \cdot \frac {1}{x},$$ $$y = \frac {x ^ {3}}{4} + \frac {C}{x}.$$

Answer: $y = \frac{x^3}{4} + \frac{C}{x}$.

Answer provided by https://www.AssignmentExpert.com