Solve the differential equation by variation of parameters. y''+3y'+2y = cos(e^x)

Solution:

We begin with the associated homogeneous equation :

y''+3y'+2y = 0

The solution of which is:

\( y_{h} = c_{1}e^{-x} + c_{2}e^{-2x} \)

Next we find a particular solution using variation of parameters method:

y₁ = \( e^{-x} \)

y₂ = \( e^{-2x} \)

g(x) = \( cos(e^{x}) \)

W(y₁, y₂) = y₁y₂’ - y₁’y₂

= \( e^{-x}(-2)( e^{-2x}) - (-( e^{-x})(e^{-2x}) \)

=-2( \( e^{-3x}) + e^{-3x} \)

=- \( e^{-3x} \)

u₁ = - \( \int \frac{y_{2}g(x)}{W(y_{1},y_{2})}dx \)

= - \( \int \frac{e^{-2x}Cos(e^{x})}{-e^{-3x}} \)

= \( \int e^{x}Cos(e^{x}) \)

= \( sin(e^{x}) \)

u₂ = \( \int \frac{y_{1}g(x))}{W(y_{1},y_{2})}dx \)

= \( \int \frac{e^{-x}Cos(e^{-x})}{-e^{-3x}} \)

= \( \int e^{2x}Cos(e^{x})dx \)

= - \( e^{x}Sin(e^{x}) - Cos(e^{x}) \)

\( y_{p} = y_{1}u_{1}+ y_{2}u_{2} \)

= \( e^{-x}Sin(e^{x}) + e^{-2x}(-e^{x}(Sin(e^{x}) - Cos(e^{x})) \)

= -\( -e^{-2x}Cos(e^{x}) \)

y = \( y_{h} + y_{p} \)

y = \( c_{1}e^{-x} + c_{2}e^{-2x} -e^{-2x}Cos(e^{x})\)

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Solve the differential equation by variation of parameters. y''+3y'+2y = cos(e^x)

Summary:

Using the variation of parameters method the solution of the equation is y = \( c_{1}e^{-x} + c_{2}e^{-2x} -e^{-2x}Cos(e^{x})\)