Question

Given that \({y_p} = \sin \left( {{x^2}} \right)\) is a particular solution to \({y^{(4)}} + y = \left( {16{x^4} - 11} \right)\sin \left( {{x^2}} \right) - 48{x^2}\cos \left( {{x^2}} \right)\)on\((0,\infty )\), find a general solution.

Short answer

The general solution is:

\(\begin{array}{c}y(x) = {y_h}(x) + {y_p}(x)y(x) = {c_1}{e^{ - \frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_2}{e^{ - \frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x\\ + {c_3}{e^{\frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_4}{e^{\frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + \sin \left( {{x^2}} \right)\end{array}\)

Step-by-step solution

Determine the general solution to the given homogeneous equation.

Consider the given differential equation;

\({y^{(4)}} + y = \left( {16{x^2} - 11} \right){\sin ^2}\left( {{x^2}} \right) - 48{x^2}\cos \left( {{x^2}} \right)\)

A particular solution to the given equation on \((0,\infty )\) is;

\({y_p}(x) = \sin \left( {{x^2}} \right).\)

Now let’s solve the corresponding homogeneous equation;

\({y^{(4)}} + y = 0\)

The auxiliary equation is:

\(\begin{array}{*{20}{r}}{{r^4} + 1 = 0}\\{{r^4} + 2{r^2} + 1 - 2{r^2} = 0}\\{{{\left( {{r^2} + 1} \right)}^2} - {{(\sqrt 2 r)}^2} = 0}\\{\left( {{r^2} + 1 + \sqrt 2 r} \right)\left( {{r^2} + 1 - \sqrt 2 r} \right) = 0}\\{\left( {{r^2} + \sqrt 2 r + 1} \right)\left( {{r^2} - \sqrt 2 r + 1} \right) = 0}\end{array}\)

Now we solve \({r^2} - \sqrt 2 r + 1 = 0\)

\(\begin{array}{c}r = \frac{{ - \sqrt 2 \pm \sqrt {2 - 4} }}{2}\\ = \frac{{ - \sqrt 2 \pm \sqrt 2 i}}{2}\\ = - \frac{{\sqrt 2 }}{2} \pm \frac{{\sqrt 2 }}{2}i\end{array}\)

Similarly solving \({r^2} + \sqrt 2 r + 1 = 0\) we get the complete set of solutions of the auxiliary equation.

\(\left\{ { - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i, - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i,\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i,\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \right\}\)

Hence, a general solution to the corresponding homogeneous equation is:

\({y_h}(x) = {c_1}{e^{ - \frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_2}{e^{ - \frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + {c_3}{e^{\frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_4}{e^{\frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x\)

Determine the general solution to the given non-homogeneous equation.

Finally, a general solution to the given non-homogeneous equation is:

\(\begin{array}{r}y(x) = {y_h}(x) + {y_p}(x)y(x) = {c_1}{e^{ - \frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_2}{e^{ - \frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + {c_3}{e^{\frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x\\ + {c_4}{e^{\frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + \sin \left( {{x^2}} \right)\end{array}\)

Thus, the general solution is:

\(y(x) = {y_h}(x) + {y_p}(x)y(x) = {c_1}{e^{ - \frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_2}{e^{ - \frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + {c_3}{e^{\frac{{\sqrt 2 }}{2}x}}\cos \frac{{\sqrt 2 }}{2}x + {c_4}{e^{\frac{{\sqrt 2 }}{2}x}}\sin \frac{{\sqrt 2 }}{2}x + \sin \left( {{x^2}} \right)\)