The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find
a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation. y'' − 7y' + 6y = x; y1 = ex
1 answer:
Answer:
y2 = (6x + 7)/36 + (Dx + E)e^x
Step-by-step explanation:
The method of reduction of order is applicable for second-order differential equations.
For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').
The differential equation becomes easy to solve, and all that is left are integration and substitutions.
Check attachments for the solution to this problem.
You might be interested in
Answer:
see the explanation
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠PQS+m∠SQR=m∠PQR ----> equation A (by Addition Angle Postulate)
we have that
m∠PQR=90° ----> equation B given problem (because is a right angle)
substitute equation B in equation A
m∠PQS+m∠SQR=90°
Remember that
Two angles re complementary is their sum is equal to 90 degrees (Definition of complementary angles)
therefore
m∠PQS and m∠SQR are complementary angles
