Question

Solve this 2nd Order Cauchy-Euler non-homogeneous differential Equation by substitution method using (x = e^t) x^2y'' - xy' + y = lnx ?

Solve this 2nd Order Cauchy-Euler non-homogeneous differential Equation by substitution method using `(x = e^t):color(white)("d")`

`x^2y'' - xy' + y = lnx` ?

Answer 1

`y=Lnx+2+C_1xLnx+C_2x`

Explanation:

After using `x=e^t` transform, `dx=e^t*dt` or `dt/dx=1/e^t=e^(-t)`

After using chain rule,

`y'=dy/dx`

=`dy/dt*dt/dx`

=`dy/dt*e^(-t)` and,

`y''=dy/dt(dy/dt*e^(-t))*dt/dx`

=`(d^2y)/dt^2*e^(-2t)-dy/dt*e^(-2t)`

Hence,

`x^2*y''-xy'+y=Lnx`

`e^(2t)*[(d^2y)/dt^2*e^(-2t)-dy/dt*e^(-2t)]-e^t*dy/dt*e^(-t)+y=Ln(e^t)`

`(d^2y)/dt^2-2dy/dt+y=t`

After setting `dy/dt=Dy` and `(d^2y)/dt^2=D^2y`

`D^2y-2Dy+y=t`

`(D^2-2D+1)*y=t`

`(D-1)^2*y=t`

`(D-1)^2*y*e^(-t)=te^(-t)`

`[(D-1)*y*e^(-t)]'=te^(-t)`

`(D-1)*y*e^(-t)=int te^(-t)*dt`

`(D-1)*y*e^(-t)=-te^(-t)-e^(-t)+C_1`

`[y*e^(-t)]'=-te^(-t)-e^(-t)+C_1`

`y*e^(-t)=-int te^(-t)*dt-int e^(-t)*dt+C_1int dt`

`y*e^(-t)=-[-te^(-t)-e^(-t)]+e^(-t)+C_1t+C_2`

`y*e^(-t)=te^(-t)+2e^(-t)+C_1t+C_2`

`y=t+2+C_1te^t+C_2e^t`

After using `x=e^t` and `t=lnx` inverse transforms, I found

`y=Lnx+2+C_1xLnx+C_2x`