Question

Find the particular solution of dy/dx+2y=2x2+3

Answer 1

Answer on Question #69647 – Math – Differential Equations

Question

Find a particular solution of

$$\frac{dy}{dx} + 2y = 2x^2 + 3.$$

Solution

Method 1

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

$$\frac{dy}{dx} + P(x)y = Q(x)$$

with $P(x) = 2$ and $Q(x) = 2x^2 + 3$.

The associated homogeneous differential equation is

$$\frac{dy}{dx} + P(x)y = 0$$ $$\frac{dy}{dx} + 2y = 0$$

The characteristic equation is

$$\lambda + 2 = 0.$$

Its root is

$$\lambda = -2.$$

The solution of the homogeneous equation is

$$y_h = c_1 e^{-2x}$$

The right-hand side of the differential equation

$$\frac{dy}{dx} + 2y = 2x^2 + 3$$

is a second degree polynomial that does not coincide with (1), therefore, a particular solution of (2) can be found in the following way:

$$y_p = a x^2 + b x + c,$$ $$y_p' = 2a x + b.$$

Substituting formulae (3), (4) into the differential equation (2) one gets

$$\frac{dy_p}{dx} + 2y_p = 2x^2 + 3$$ $$2a x + b + 2(a x^2 + b x + c) = 2x^2 + 3$$ $$2a x^2 + (2b + 2a) x + (b + 2c) = 2x^2 + 3,$$

Equating coefficients before the like terms one gets the system

$$\left\{ \begin{array}{c} 2a = 2 \\ 2b + 2a = 0 \\ b + 2c = 3 \end{array} \right.$$ $$\left\{ \begin{array}{c} a = 1 \\ b = -a \\ c = \frac{3 - b}{2} \end{array} \right.$$ $$\left\{ \begin{array}{l} a = 1 \\ b = -1 \\ c = 2 \end{array} \right.$$

Thus, a particular solution of (2) is

$$y_p = a x^2 + b x + c = x^2 - x + 2$$

Method 2

A particular solution can be found by means of the general solution of the differential equation.

**The first way**

is application of an auxiliary function.

Let

$$y = u v,$$

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

Then

$$\frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx}.$$

So we have

$$\frac{du}{dx} v + u \frac{dv}{dx} + 2u v = 2x^2 + 3$$

or

$$\frac{du}{dx} v + u \left(\frac{dv}{dx} + 2v\right) = 2x^2 + 3.$$

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

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

So we have

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

or

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

It’s a separable equation which can be easily integrated:

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

Substitute the expression for the function $v$ into equation (1):

$$\frac {du}{dx} e^{-2x} = 2x^2 + 3$$

or

$$\frac {du}{dx} = (2x^2 + 3) e^{2x}.$$

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

$$\int du = \int (2x^2 + 3) e^{2x} dx.$$

To calculate the second integral we shall use the formula for integration by parts:

$$\int udv = uv - \int v du.$$

Let

$$u = 2x^2 + 3 \quad \text{and} \quad dv = e^{2x} dx,$$

then

$$du = 4x dx; \quad v = \frac{1}{2} e^{2x}.$$

Using the formula for integration by parts, we have

$$\int (2x^2 + 3) e^{2x} dx = \frac{(2x^2 + 3)}{2} e^{2x} - 2 \int x e^{2x} dx.$$

This integral still contains a product of functions. So we shall use the formula for integration by parts again. This time we choose

$$u = x \quad \text{and} \quad dv = e^{2x} dx$$

then

$$du = dx; \quad v = \frac{1}{2} e^{2x}.$$

So

$$\begin{aligned} & \int (2x^2 + 3) e^{2x} dx = \frac{(2x^2 + 3)}{2} e^{2x} - 2 \int x e^{2x} dx = \\ & = \frac{(2x^2 + 3)}{2} e^{2x} - 2 \left( \frac{x}{2} e^{2x} - \frac{1}{2} \int e^{2x} dx \right) = \frac{(2x^2 + 3)}{2} e^{2x} - x e^{2x} + \frac{1}{2} e^{2x} + C = \\ & = (x^2 - x + 2) e^{2x} + C, \end{aligned}$$

where $C$ is an integration constant.

So we get the following auxiliary function:

$$u = (x ^ {2} - x + 2) e ^ {2 x} + C.$$

After the back substitution we have the general solution of the given equation:

$$y = x ^ {2} - x + 2 + C e ^ {- 2 x},$$

where $C$ is arbitrary real constant.

A particular solution can be obtained if in (2) we set $C$ to be equal to a certain number. For example, if $C = 0$, then $y_{p} = x^{2} - x + 2$ will be a particular solution.

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

$$y = e ^ {- 2 x} \int (2 x ^ {2} + 3) e ^ {2 x} d x = e ^ {- 2 x} \big ((x ^ {2} - x + 2) e ^ {2 x} + C \big) = x ^ {2} - x + 2 + C e ^ {- 2 x},$$

where $C$ is an integration constant.

So the general solution of the given equation is

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

A particular solution can be obtained if in (3) we set $C$ to be equal to a certain number. For example, if $C = 0$, then $y_{p} = x^{2} - x + 2$ will be a particular solution.

**Answer**: $y_{p} = x^{2} - x + 2$.

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