Question

Solve the diferential equation y′=−y+x2y2

Answer 1

Answer on Question #69646, Math / Differential Equations

Solve the differential equation $y' = -y + x^2 y^2$

Solution

It's not hard to see that this is a Bernoulli differential equation

$$y' + p(x)y = q(x)y^n$$

In this case, we have that $n = 2$

$$y' + y = x^2 y^2.$$

Then to find the solution, change the dependent variable from $y$ to $z$,

where $z = y^{-1}$. This gives a differential equation in $x$ and $z$ that is linear, and can be solved using the integrating factor method.

$$\begin{array}{l} y = \frac{1}{z}, y' = \left(\frac{1}{z}\right)' = -\frac{1}{z^2} z' \\ \quad -\frac{1}{z^2} z' + \frac{1}{z} = x^2 \frac{1}{z^2} \\ \quad -z' + z = x^2 \\ \quad z' - z = -x^2 \\ \end{array}$$

Integrating Factor $= v(x) = e^{\int p(x)dx} = e^{\int (-1)dx} = e^{-x}$

Constant of integration is 0, so $v$ is simple as possible.

Multiple both sides by a positive function $v(x)$ that transforms the left-hand side into the derivative of the product $v(x) \cdot z$.

$$e^{-x} \frac{dz}{dx} - e^{-x}z = -x^2 e^{-x}$$ $$\begin{array}{l} \frac{d}{dx}(e^{-x}z) = e^{-x} \frac{dz}{dx} - e^{-x}z \\ \frac{d}{dx}(e^{-x}z) = -x^2 e^{-x} \\ \int d(e^{-x}z) = \int (-x^2 e^{-x}) dx \\ \end{array}$$ $$\begin{array}{l} \int (-x^2 e^{-x}) dx = x^2 e^{-x} - 2 \int x e^{-x} dx = x^2 e^{-x} + 2x e^{-x} + 2 e^{-x} + C \\ \int u dv = uv - \int v du \\ u = x^2, du = 2xdx \\ dv = -e^{-x}dx, v = e^{-x} \\ \quad -2 \int x e^{-x} dx = 2x e^{-x} - 2 \int e^{-x} dx = 2x e^{-x} + 2 e^{-x} + C \\ \int u dv = uv - \int v du \\ u = 2x, du = 2dx \\ dv = -e^{-x}dx, v = e^{-x} \\ \end{array}$$ $$\begin{array}{l} e^{-x} z = x^{2} e^{-x} + 2x e^{-x} + 2e^{-x} + C \\ z = x^{2} + 2x + 2 + C e^{x} \\ \text{Put back } z = y^{-1} \\ y = \frac{1}{x^{2} + 2x + 2 + C e^{x}} \\ \end{array}$$

Answer: $y = \frac{1}{x^{2} + 2x + 2 + C e^{x}}$.

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