Question

Find y as a function of x if rmula1" title=" x^{2} y''-4xy'-6y= x^{3} " alt=" x^{2} y''-4xy'-6y= x^{3} " align="absmiddle" class="latex-formula">
y(1)=6 and y'(1)=-1

Answer 1

Let $y(x)=z(t)$, where $t=\ln x$. Then

$\implies y'=\dfrac1xz'$
$\implies y''=\dfrac1{x^2}(z''-z')$

so the ODE is equivalent to

$(z''-z')-4z'-6z=e^{3t}$
$z''-5z'-6z=e^{3t}$

The characteristic equation is

$r^2-5r-6=(r-6)(r+1)=0\implies r=6,r=-1$

so that the characteristic solution is

$z_c=C_1e^{6t}+C_2e^{-t}$

For the particular solution, take

$z_p=ae^{3t}$
$\implies {z_p}'=3ae^{3t}$
$\implies {z_p}''=9ae^{3t}$

$\implies 9ae^{3t}-15ae^{3t}-6ae^{3t}=e^{3t}$
$\implies -12a=1$
$\implies a=-\dfrac1{12}$

$\implies z_p=-\dfrac1{12}e^{3t}$

$\implies y_p=-\dfrac1{12}x^3$

$\implies y=y_c+y_p=C_1x^6+\dfrac{C_2}x-\dfrac1{12}x^3$

With the initial conditions, we get

$y(1)=6\implies 6=C_1+C_2-\dfrac1{12}$
$y'(1)=-1\implies -1=6C_1-C_2-\dfrac14$
$\implies C_1=\dfrac{16}{21},C_2=\dfrac{149}{28}$

So the particular solution to the IVP is

$y=\dfrac{16x^6}{21}+\dfrac{149}{28x}-\dfrac{x^3}{12}$