y\"-2y\'-3y=xsin(pi*x) A) Find the solution to the homogenous equation, that is,
ID: 3419906 • Letter: Y
Question
y"-2y'-3y=xsin(pi*x) A) Find the solution to the homogenous equation, that is, find yh(x) B) what is R(x)? C) find the form of a particular solution D) Find y'p(x) and y"p(x) E) find yp(x) F) find the general solution to the differential equation. 2. Given y" + 2y,-3y = x sin(nz) (a) (3 points) Find the solution to the homogeneous equation, that is, find sx (a) (o) (3 points) Fnd the solution to the homogeneous equation, that is, find g,(a) (b) (1 point) What is R(x)? (c) (2 points) Find the form of a particular solution (d) (4 points) Finds(x) and (z)Explanation / Answer
Post one more question to get the answers to the remaining part. Thanks
a) In order to solve for homogeneous equation, we need to solve for LHS equation equating it to zero
y'' + 2y' - 3y =0
we need to write the characteristic equation that comes with p^2 coefficient with y'', p coefficient with y' term and 1 coefficient with y term
Hence Characteristic equation will be come
p^2 + 2p - 3 = 0
p^2 - p + 3p - 3 = 0
p(p-1) + 3(p-1) = 0
(p-1)(p+3) = 0
Hence p=1,-3
Homogeneous solution is
yh(x) = c1*e^(x) + c2 * e^(-3x)
b) The general second order equation is of the form
y'' + p(x)y' + q(x)y = r(x)
Hence R(x) = x*sin(pi*x) in the above equation
c) In the third question we need to write the form of yp
since the right hand side terms contains the multiplication of polynomial with the sine term, the function will be similar to having particular solution
yp = (Ax^2 + Bx + C)(Dsin(pi*x) + Ecos(pi*x))
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