Question

Find the general solution of the differential equation using the method of undetermined coefficients y" + y' - 6y = x^2

Answer 1

First we find the solution of the homogeneous part y" + y - 6y = 0

Let y = e^mx be a solution of the above equation then ,

m² + m - 6 = 0

Or , (m + 3 ) (m - 2 ) =0

Or , m = 2 , -3

So the solution of homogeneous part is ,

y (x) = c₁ e^-3x + c₂ e^2x , where c₁ , care are arbitrary constants .

Let's find the complementary part using method of undetermined coefficient .

Let y_p = ax² + bx + c where a , b ,c are constants whose value we have to determine .

y_p^' = 2ax + b

y_p" = 2a

Substituting the value of y_p , y_p^' , y_p^" in the given equation we get ,

2a + 2ax + b - 6( ax² + bx + c ) = x²

Or , -6ax² + (2a - 6b ) + 2a + b - 6c = x²

Equating coefficient both side we get ,

-6a = 1 ,

2a - 6b = 0

2a + b - 6c = 0

So ,

Hence the required solution is ,

y(x) = c₁ e^3x + c₂e^2x .

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