Question

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

Answer 1

$$(x-1)u''-xu'+u=0, \;x>1$$

$u=a_1u_1+a_2u_2$ 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

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

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

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

Answer: a)$u_2=xe^x$