# Dy/dx = x+2y How to solve this?

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

The General Solution is:

# y = -1/2x -1/4 + Ce^(2x) #

#### Explanation:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

We have:

# dy/dx = x+2y #

Which we can write as:

# dy/dx -2y = x # ..... [A]

This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor,

# I = e^(int P(x) dx) #

# \ \ = exp(int \ -2 \ dx) #

# \ \ = exp( -2x) #

# \ \ = e^(-2x) #

And if we multiply the DE [A] by this Integrating Factor,

# dy/dxe^(-2x) -2ye^(-2x) = xe^(-2x) #

# :. d/dx( ye^(-2x) ) = xe^(-2x) #

This has transformed our initial ODE into a Separable ODE, so we can now *"separate the variables"* to get::

# ye^(-2x) = int \ xe^(-2x) \ dx #

We can proceed via an application of integration by Parts

Let

# { (u,=x, => (du)/dx,=1), ((dv)/dx,=e^(-2x), => v,=-1/2e^(-2x) ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

We have:

# int \ (x)(e^(-2x)) \ dx = (x)(-1/2e^(-2x)) - int \ (-1/2e^(-2x))(1) \ dx #

# :. int \ xe^(-2x) \ dx = -1/2xe^(-2x) +1/2 \ int \ e^(-2x) \ dx #

# " " = -1/2xe^(-2x) -1/4 e^(-2x) + C #

Using this result, we can write the DE Solution as :

# ye^(-2x) = -1/2xe^(-2x) -1/4 e^(-2x) + C #

# :. y = -1/2x -1/4 + Ce^(2x) #

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