Find the form of a particular solution to the differential equations below. I do not attempt to find the coefficients. y''+2y'= 5x^2 + x + x^3e^(-2x)+cos(2x)

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embizze | High School Teacher | (Level 1) Educator Emeritus

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The general solution is `y=c_1e^(-2x)+c_2` .

(1)The particular solution `p(x)` will be the sum of the particular solutions:

`p_1(x)` is the part. sol. to `y''+2y'=5x^2+x`

`p_2(x)` is the part. sol. to `y''+2y'=e^(-2x)x^3`

`p_3(x)` is the part. sol. to `y''+2y'=cos(2x)`

(2) `p_1(x)=x(a_1+a_2x+a_3x^2)` Note that we multiply by x to account for the constant term `c_2` in the general solution.

(3) `p_2(x)=x(a_4e^(-2x)+a_5xe^(-2x)+a_6x^2e^(-2x)+a_7x^3e^(-2x))`

` ` Note that again we multiply by x to account for `e^(-2x)` in the general solution.

(4) `p_3(x)=a_8cos(2x)+a_9sin(2x)`

(5) Thus the particular solution has the form:

`p(x)=x(a_1+a_2x+a_3x^2)+x(a_4e^(-2x)+a_5xe^(-2x)+a_6x^2e^(-2x)+a_7x^3e^(-2x))+a_8cos(2x)+a_9sin(2x)`

` ` To solve for the coefficients, take `p'(x),p''(x)` ; plug into the differential equation and equate coefficients. This is messy, but not too bad.

The solution is:

`-182e^(-2x)x^4+(5/6-1/4e^(-2x))x^3+(-1-3/8e^(-2x))x^2+(1-3/8e^(-2x))x-1/8(cos(2x)sin(2x))+c_1e^(-2x)+c_2`

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