練習問題1
\(y’=x^2 y\ \)を解け。
[解答]
$$y’=x^2y$$ $$\displaystyle\frac{dy}{dx}=x^2y$$ $$\displaystyle\frac{1}{y}\displaystyle\frac{dy}{dx}=x^2$$ $$\displaystyle\frac{1}{y}dy=x^2dx$$ $$\int\displaystyle\frac{1}{y}dy=\int{x^2dx}$$ $$\log{|y|}=\displaystyle\frac{1}{3}x^3+C_1\\\left(C_1は積分定数\right)$$ $$y=\mathrm{e}^{\frac{1}{3}x^3+C_1}$$ $$y=C\mathrm{e}^{\frac{1}{3}x^3}\\\left(綺麗にするためC=\mathrm{e}^{C_1}と置き換え\right)$$
練習問題2
\(y’=4y^\frac{3}{4}\ \)を解け。
[解答]
$$y’=4y^{\frac{3}{4}}$$
$$\displaystyle\frac{dy}{dx}=4y^{\frac{3}{4}}$$
$$y^{-\frac{3}{4}}\displaystyle\frac{dy}{dx}=4$$
$$y^{-\frac{3}{4}}dy=4dx$$
$$\int{y^{-\frac{3}{4}}dy}=\int4dx$$
$$\displaystyle\frac{1}{-\frac{3}{4}+1}y^{-\frac{3}{4}+1}=4x+C_1\\\left(C_1は積分定数\right)$$
$$4y^{\frac{1}{4}}=4x+C_1$$
$$y^{\frac{1}{4}}=x+C\\\left(\displaystyle\frac{C_1}{4}=Cと置き換え\right)$$
$$y=\left(x+C\right)^4$$
練習問題3
\((1+x^2)y’=y\ \)を解け。
$$\left(1+x^2\right)y’=y$$
$$y’=\displaystyle\frac{y}{1+x^2}$$
$$\displaystyle\frac{dy}{dx}=\displaystyle\frac{y}{1+x^2}$$
$$\displaystyle\frac{1}{y}\displaystyle\frac{dy}{dx}=\displaystyle\frac{1}{1+x^2}$$
$$\int\displaystyle\frac{1}{y}dy=\int\displaystyle\frac{1}{1+x^2}dx$$
$$\log{|y|}=\tan{x}+C_1\\\left(C_1は積分定数\right)$$
$$y=\mathrm{e}^{\tan{x}+C_1}$$
$$y=C\mathrm{e}^{\tan{x}}\\\left(C=\mathrm{e}^{C_1}\right)$$
練習問題4
\(y’=y\cos {x}\ \)を解け。
[解答]
$$y’=y\cos{x}$$ $$\displaystyle\frac{dy}{dx}=y\cos{x}$$ $$\displaystyle\frac{1}{y}\displaystyle\frac{dy}{dx}=\cos{x}$$ $$\displaystyle\frac{1}{y}dy=\cos{x}dx$$ $$\int\displaystyle\frac{1}{y}dy=\int\cos{x}dx$$ $$\log{|y|}=\sin{x}+C_1$$ $$y=\mathrm{e}^{\sin{x}+C_1}\\\left(C_1は積分定数\right)$$ $$y=C\mathrm{e}^{\sin{x}}\\\left(C=\mathrm{e}^{C_1}\right)$$
練習問題5
\(yy’\sqrt{1-x^2}=1\ \)を解け。
[解答]
$$yy’\sqrt{1-x^2}=1\ $$ $$y’=\displaystyle\frac{1}{y\sqrt{1-x^2}}$$ $$\displaystyle\frac{dy}{dx}=\displaystyle\frac{1}{y\sqrt{1-x^2}}$$ $$y\displaystyle\frac{dy}{dx}=\displaystyle\frac{1}{\sqrt{1-x^2}}$$ $$ydy=\displaystyle\frac{1}{\sqrt{1-x^2}}dx$$ $$\int{ydy}=\int\displaystyle\frac{1}{\sqrt{1-x^2}}dx\cdot\cdot\cdot\left(A\right)$$ \(\left(A\right)\)の右辺について, \(x=\sin{\theta}\) と置換すると \(dx=\cos{\theta}d\theta\)だから $$\int\displaystyle\frac{1}{\sqrt{1-x^2}}dx=\int\displaystyle\frac{1}{\sqrt{1-\sin^2{\theta}}}\cdot\cos{\theta}d\theta$$ \(\sqrt{1-\sin^2{\theta}}=\cos{\theta}\)より \begin{eqnarray}\int\displaystyle\frac{1}{\sqrt{1-x^2}}dx&=&\int\displaystyle\frac{1}{\cos{\theta}}\cdot\cos{\theta}d\theta\\&=&\int{d\theta}\\&=&\theta+C_1\\&=& \sin^{-1}{x}+C_1\end{eqnarray} 従って, \(\left(A\right)\)式は $$\displaystyle\frac{1}{2}y^2=\sin^{-1}{x}+C_1$$ $$\displaystyle\frac{1}{2}y^2+C=\sin^{-1}{x}\\\left(C=-C_1\right)$$ $$\sin{\left(\displaystyle\frac{1}{2}y^2+C\right)}=x$$
