Answer to Question #91482 in Differential Equations for Sajid

Question #91482
Q.Choose the correct answer.
If one of the solution of the differential equation (x-1)u’’-xu’+u=0, x>1 is
U=ex, then which of the following is the second linearly independent solution?
a)u=xex
b) u=3ex
c)u=x
d)u=x2+1
1
Expert's answer
2019-07-10T11:48:29-0400
"(x-1)u''-xu'+u=0, \\;x>1"

"u=a_1u_1+a_2u_2" is a solution of the differential equation

"u_1=e^x" is the first solution.

Make a substitution "u=u_1\\int y(x)dx=e^x\\int y(x)dx \\;\\;\\;\\;\\;\\;(1)"

"\\int y(x) = \\frac{u}{e^x}"

then


"u'=(e^x \\int y(x)dx)'=e^x\\int y(x)dx + e^x(\\int y(x)dx)'="

"=e^x \\int y(x)dx+e^xy(x)=e^x\\frac{u}{e^x}+e^xy(x)=u+e^xy(x)"

"u'' = (e^x\\int y(x)dx +e^xy(x))'="

"=u+e^xy(x)+e^xy(x)+e^xy'(x)=u+2e^xy(x)+e^xy'(x)"

then


"(x-1)(u+2e^xy+e^xy')-x(u+e^xy)+u=0 \\iff"

"xu+2xe^xy+xe^xy'-u-2e^xy-e^xy'-xu-xe^xy+u=0 \\iff"

"e^x(x-1)y'+e^x(x-2)y=0 \\iff"

"(x-1)y'=(2-x)y \\iff""(x-1)\\frac{dy}{dx}=(2-x)y \\iff"

"\\frac{1}{y}dy=\\frac{2-x}{x-1}dx"

"\\int \\frac{1}{y}dy=\\int (\\frac{1}{x-1} -1)dx"

"ln|y|+C_1 = ln|x-1|-x+C_2 \\implies"

"y = C(x-1)e^x \\;\\;\\;\\;\\;\\;\\; (2)"

From (1) and (2) we get


"u=Ce^x\\int (x-1)e^xdx=\\begin{vmatrix}\n u=x-1 & dv=e^xdx \\\\\n du=dx & v=e^x\n\\end{vmatrix} ="

"=C(e^x(x-1)-\\int e^xdx)=C(e^x(x-1)-e^x+C_3)=C(xe^x-2e^x+C_3)"


Then "u_1=e^x" is the first solution and "u_2=xe^x" is the second linearly independent solution.

Answer: a)"u_2=xe^x"


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