Author Topic: Q4-T0601  (Read 3591 times)

Victor Ivrii

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« on: March 02, 2018, 05:28:05 PM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
&x^2y'' + xy' + (x^2 - 0.25)y = 3x^{3/2} \sin (x), &&x > 0;\\
&y_1(x) = x^{-1/2} \sin (x),\qquad y_2(x) = x^{-1/2} \cos (x)
« Last Edit: March 02, 2018, 05:31:58 PM by Victor Ivrii »

Meng Wu

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Re: Q4-T0601
« Reply #1 on: March 02, 2018, 05:34:10 PM »
$$x^2y''+xy'+(x^2-0.25)y=3x^{3\over2}\sin(x); x>0; \\ y_1(x)=x^{-{1\over2}}\sin(x), y_2(x)=x^{-{1\over2}}\cos(x)$$
Hence,$$\cases{y_1(x)=x^{-{1\over2}}\sin x\\y_1'(x)=-{1\over2}x^{-{3\over 2}}\sin x+x^{-{1\over2}}\cos x\\y_1’’(x)={3\over4}x^{-{5\over2}}\sin x-x^{-{3\over2}}\cos x-x^{-{1\over2}}\sin x}$$
$$\cases{y_2(x)=x^{-{1\over2}}\cos(x)\\y_2’(x)=-{1\over2}x^{-{3\over2}}\cos x-x^{-{1\over2}}\sin x \\ y_2’’(x)={3\over4}x^{-{5\over2}}\cos x+x^{-{3\over2}}\sin x-x^{-{1\over2}}\cos x}$$
Substitute back into the homogeneous equation: $$x^2y''+xy'+(x^2-0.25)y=0$$
Verified that $y_1(x)$ and $y_2(x)$ both satisfy the corresponding homogeneous equation. $\\$
And the complementary solution $y_c(x)=c_1x^{-{1\over2}}\sin x+c_2x^{-{1\over2}}\cos x$ $\\$
Now divide both sides of the original equation by $x^2$:
$$y’’+{1\over x}y'+{x^2-0.25\over x^2}y=3x^{-{1\over2}}\sin x$$
Then $$p(t)={1\over x}, q(t)={x^2-0.25\over x^2}, g(t)=3x^{-{1\over2}}\sin x$$
$$W[y_1,y_2](x)=\begin{array}{|c c|} y_1(x)&y_2(x)\\y_1’(x)&y_2’(x)\end{array}=-x^{-1}\neq 0$$
Since the particular solution has the form: $$Y(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$$
and $$\begin{align}u_1(x)&=-\int{{y_2(x)g(x)\over W[y_1,y_2](x)}}dx\\&=-\int{x^{-{1\over2}}\cos x\cdot 3x^{-{1\over2}}\sin x\over -x^{-1}}dx\\&={3\over2}\int{2\sin x\cos x}dx\\&={3\over2}\int{\sin 2x}dx\\&=-{3\over4}\cos 2x\end{align}$$
$$\begin{align}u_2(x)&=\int{y_1(x)g(x)\over W[y_1,y_2](x)}dx\\&=\int{x^{-{1\over2}}\sin x\cdot 3x^{-{1\over2}}\sin x\over -x^{-1}}dx\\&=-3\int{\sin^2 x}dx\\&=-3\int{1-\cos 2x\over 2}dx\\&=-{3\over2}x+{3\over4}\sin 2x\end{align}$$
Therefore, $$\begin{align}Y(x)&=-{3\over4}\cos 2x\cdot x^{-{1\over2}}\sin x+ (-{3\over2}x+{3\over4}\sin 2x)\cdot x^{-{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}(\sin 2x\cos x-\cos 2x\sin x)-{3\over 2}x^{-{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}(2\sin x\cos^2 x-(2\cos^2 x-\cos x)\sin x)-{3\over 2}x^{{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}\sin x-{3\over 2}x^{{1\over2}}\cos x\end{align}$$
Hence, the general solution:
$$\begin{align}y(x)&=y_c(x)+Y(x)\\&=c_1x^{-{1\over2}}\sin x +c_2x^{-{1\over2}}\cos x+ {3\over4}x^{-{1\over2}}\sin x-{3\over 2}x^{{1\over2}}\cos x\\&= (c_1+{3\over4})x^{-{1\over2}}\sin x+ c_2x^{-{1\over2}}\cos x-{3\over2}x^{1\over2}\cos x \\&=c_3x^{-{1\over2}}\sin x+ c_2x^{-{1\over2}}\cos x-{3\over2}x^{1\over2}\cos x\end{align}$$
where $c_3=c_1+{3\over4}$ $\\$
Therefore, the particular solution of the given nonhomogeneous equation is $$Y_{new}(t)=-{3\over2}x^{1\over2}\cos x$$