Answer to Question #154890 in Differential Equations for berkay

Question #154890

y''(x) - y'(x) + 2y(x) = 2e^x


1
Expert's answer
2021-01-14T17:54:54-0500

Let's solve this problem

"y''\\:-y'\\:+2y=2e^x:"

Second-order linear non-homogeneous differential equation with constant coefficients

A second order linear, non-homogeneous ODE has formed of "ay''+by'+cy=g\\left(x\\right)"

The General solution to "a\\left(x\\right)y''+b\\left(x\\right)y'+c\\left(x\\right)y=g\\left(x\\right)" can be written as

"y=y_h+y_p"

"y_h" is the solution to the homogeneous ODE "a\\left(x\\right)y''+b\\left(x\\right)y'+c\\left(x\\right)y=0"

"y_p" the particular solution, is any function That satisfies the non-homogenous equation

Find "y_h" by solving "y''\\:-y'\\:+2y = 0"; "y=e^{\\frac{x}{2}}\\left(c_1\\cos \\left(\\frac{\\sqrt{7}x}{2}\\right)+c_2\\sin \\left(\\frac{\\sqrt{7}x}{2}\\right)\\right)"

Find "y_p" by solving "y''\\:-y'\\:+2y=2e^x:" "y = e^x"

The general solution "y=y_h+y_p"

"y=e^{\\frac{x}{2}}\\left(c_1\\cos \\left(\\frac{\\sqrt{7}x}{2}\\right)+c_2\\sin \\left(\\frac{\\sqrt{7}x}{2}\\right)\\right)+e^x"


"\\mathrm{Plotting:}\\:e^{\\frac{x}{2}}\\left(c_1\\cos \\left(\\frac{\\sqrt{7}x}{2}\\right)+c_2\\sin \\left(\\frac{\\sqrt{7}x}{2}\\right)\\right)+e^x\\quad \\mathrm{assuming}\\quad \\:c_1=1\\quad \\:c_2=1"


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