For a linear, inhomogenous ordinary differential equation: <mstyle displaystyle="true" scriptlev

glycleWogry

glycleWogry

Answered question

2022-06-12

For a linear, inhomogenous ordinary differential equation:
d y d x + 2 y = 4 x 2 + 6 x 1
(a) Find one solution to equation above that is quadratic. ie. of the form P ( x ) = A x 2 + B x + C
I used the integrating factor and find out that y = 2 x 2 + x 1
(b) solve the homogeneous linear equation d y d x + 2 y = 0
And my solution is y = e 2 x + C or y = 0
(c) Solve d y d x + 2 y = 4 x 2 + 6 x 1 using (a) and (b). Hint: given a solution y to d y d x + 2 y = 4 x 2 + 6 x 1, show that y P is a solution to a nice equation.
I don't really understand part (c). Can anyone give me some hints?

Answer & Explanation

Quinn Everett

Quinn Everett

Beginner2022-06-13Added 23 answers

If L ( y ) is a linear differential operator, for example L ( y ) = y + 2 y as in your case, then every solution to L ( y ) = f can be written as
y = y h + y p
where y h is a solution to the homogeneous equation L ( y ) = 0 and y p is a particular solution to L ( y ) = f.
Proof. Let y be an arbitrary solution to L ( y ) = f. Then L ( y y p ) = L ( y ) L ( y p ) = f f = 0. Hence y y p solves the homogeneous equation.
Conversely, if y h and y p are as above, then y = y h + y p solves L ( y ) = f.
Proof. L ( y h + y p ) = L ( y h ) + L ( y p ) = 0 + f = f.
Notice the importance of L being linear.

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