東京大学新領域創成科学研究科海洋技術環境学専攻 2023年8月実施第1問

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祭音Myyura

以下の微分⽅程式を解け。

\frac{d^2y}{dx^2} + y = \sin 2x

ただし、 $x = 0$ のとき $y = \frac{dy}{dx} = 0$ 。

Solve the following differential equation:

\frac{d^2y}{dx^2} + y = \sin 2x

where $y = \frac{dy}{dx} = 0$ at $x = 0$ .

求初值问题

y''+y=\sin2x,\qquad y(0)=0,\quad y'(0)=0

的解。

y'' + y = 0.

Characteristic equation:

r^2+1=0 \quad\Rightarrow\quad r=\pm i.

So homogeneous solution:

y_h(x) = C_1 \cos x + C_2 \sin x.

Try $y_p(x) = A\sin 2x + B\cos 2x.$

y_p'' = -4A\sin 2x -4B\cos 2x.

Plug into LHS:

y_p'' + y_p = (-4A\sin 2x -4B\cos 2x) + (A\sin 2x + B\cos 2x) = (-3A)\sin 2x + (-3B)\cos 2x.

We want this to equal $\sin 2x$ . So:

-3A = 1 \quad\Rightarrow\quad A = -\tfrac{1}{3}, \quad -3B = 0 \quad\Rightarrow\quad B = 0.

Thus:

y_p(x) = -\tfrac{1}{3}\sin 2x.

y(x) = C_1 \cos x + C_2 \sin x - \tfrac{1}{3}\sin 2x.

y(0) = C_1\cdot 1 + C_2\cdot 0 - \tfrac{1}{3}\sin 0 = C_1 = 0.

y'(x) = -C_1 \sin x + C_2 \cos x - \tfrac{2}{3}\cos 2x.

y'(0) = -C_1\cdot 0 + C_2\cdot 1 - \tfrac{2}{3}\cdot 1 = C_2 - \tfrac{2}{3} = 0.

C_2 = \tfrac{2}{3}.

\boxed{y(x) = \tfrac{2}{3}\sin x - \tfrac{1}{3}\sin 2x}