Question

Find the general solution of y′+2y=x3e−2x

Answer 1

Answer on Question #69627 – Math – Differential Equations

Question

Find the general solution of

$$y' + 2y = x^3 e^{-2x}.$$

Solution

It's a first order linear differential equation, that is, it has the following form:

$$y' + P(x)y = Q(x)$$

with $P(x) = 2$ and $Q(x) = x^3 e^{-2x}$.

It can be integrated in different ways.

The first way is using auxiliary functions.

Let

$$y = uv,$$

where $u = u(x)$ and $v = v(x)$ are the auxiliary functions.

Then

$$y' = u'v + uv'.$$

So we have

$$u'v + uv' + 2uv = x^3 e^{-2x}$$

or

$$u'v + u(v' + 2v) = x^3 e^{-2x}.$$

Since one of the auxiliary functions can be chosen arbitrarily, let a function $v$ be such that it will satisfy the following condition:

$$v' + 2v = 0.$$

So we have

$$\frac{dv}{dx} + 2v = 0$$

or

$$\frac{dv}{dx} = -2v.$$

It's a separable equation that can be easily integrated:

$$\int \frac{dv}{v} = -2\int dx$$ $$\ln |v| = -2x$$ $$v = e ^ {- 2 x}.$$

Substitute the expression (4) for the function $v$ into equation (2) using (3):

$$u ^ {\prime} e ^ {- 2 x} = x ^ {3} e ^ {- 2 x}$$

or

$$u ^ {\prime} = x ^ {3}$$ $$\frac {d u}{d x} = x ^ {3}.$$

It's a separable equation so we can integrate it:

$$\int d u = \int x ^ {3} d x$$ $$u = \frac {1}{4} x ^ {4} + C,$$

where $C$ is an integration constant.

After substitution (4), (5) back into (1) we have a general solution of the given equation:

$$y = \left(\frac {1}{4} x ^ {4} + C\right) e ^ {- 2 x},$$

where $C$ is an integration constant.

The second way is using an integrating factor

$$\mu (x) = e ^ {\int p (x) d x}.$$

The solution is then commonly written as

$$y = \frac {1}{\mu (x)} \int \mu (x) Q (x) d x.$$

Determine the integrating factor:

$$\mu (x) = e ^ {2 \int d x} = e ^ {2 x}.$$

Then the solution is

$$y = e ^ {- 2 x} \int e ^ {2 x} x ^ {3} e ^ {- 2 x} d x = e ^ {- 2 x} \int x ^ {3} d x = e ^ {- 2 x} \left(\frac {1}{4} x ^ {4} + C\right).$$

So the general solution of the given equation is

$$y = \left(\frac {1}{4} x ^ {4} + C\right) e ^ {- 2 x},$$

where $C$ is an integration constant.

Answer: $y = \left(\frac{1}{4} x^4 + C\right) e^{-2x}$ .

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