常微分方程式の問題集の回答

(1)

$ \iff\mathrm dy=\mathrm d\left((x+1)^2\right)

$ \underline{\iff y=(x+1)^2+C\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \iff\mathrm d\left(y^2\right)=\mathrm d\left(2e^x\right)

$ \iff y^2=2e^x+C\quad\text{.for }\exist C\ge0

$ \underline{\iff y=\pm\sqrt{2e^x+C}\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \iff \mathrm d(x+y+1)-\mathrm dx=(x+y+1)^2\mathrm dx

$ \iff \mathrm dz-\mathrm dx=z^2\mathrm dx

$ \iff\mathrm dz=(1+z^2)\mathrm dx

$ \iff\frac1{1+z^2}\mathrm dz=\mathrm dx

$ \iff\tan^{-1}z=x+C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y=\tan(x+C)-x-1\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \iff \mathrm d(x-y)=\mathrm dx-\frac{1}{x-y}\mathrm dx

$ \iff\frac1{1-\frac1{x-y}}\mathrm d(x-y)=\mathrm dx

$ \iff\frac{x-y}{x-y-1}\mathrm d(x-y)=\mathrm dx

$ \iff\left(1+\frac{1}{(x-y)-1}\right)\mathrm d(x-y-1)=\mathrm dx

$ \iff\mathrm d(x-y-1+\ln|x-y-1|)=\mathrm x

$ \iff y+1=\ln|x-y-1|+C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff Ce^y+y=x-1\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \mathrm dy=\mathrm dx+\frac yx\mathrm dx

$ \iff x\mathrm dy-y\mathrm dx=x\mathrm dx

$ \iff\frac{x\mathrm dy-y\mathrm dx}{x^2}=\frac1x\mathrm dx

$ \iff \mathrm d\left(\frac yx\right)=\mathrm d\ln|x|

$ \iff\frac yx-\ln|x|=C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y=x(\ln|x|+C)\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

(2)

$ \iff \mathrm d(y-1)+2x(y-1)\mathrm dx=0

$ \iff\mathrm d(\ln|y-1|)+\mathrm d\left(x^2\right)=0

$ \iff\ln|y-1|+x^2=C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y=Ce^{-x^2}+1\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

もしくは

$ y'+y(x^2)'=2x

$ \iff\frac1{e^{x^2}}(ye^{x^2})'=(x^2)'

$ \iff (ye^{x^2})'=(e^{x^2})'

$ \iff ye^{x^2}=e^{x^2}+C\quad\text{.for }\exist C\in\Complex

$ \iff y=Ce^{-x^2}+1\quad\text{.for }\exist C\in\Complex

$ \iff y'+\frac1xy=\frac1{1+x^2}

$ \iff y(\ln|x|)'+y'=\frac1{1+x^2}

$ \iff\frac{1}{|x|}(y|x|)'=\frac1{1+x^2}

$ \iff\frac{1}{x}(yx)'=\frac1{1+x^2}

$ \iff 2(yx)'=(\ln|1+x^2|)'

$ \iff 2yx+C=\ln|1+x^2|\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y=\frac1{2x}(\ln|1+x^2|+C)\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

もしくは

$ xy'+y=\frac{x}{1+x^2}

$ \iff(2xy)'=(\ln|1+x^2|)'

$ \iff 2xy=\ln|1+x^2|+C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y=\frac1{2x}(\ln|1+x^2|+C)\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \iff y'+y(-\ln|\cos x|)'=2x\cos x

$ \iff|\cos x|(y/|\cos x|)'=2x\cos x

絶対値を外せるので

$ \iff\left(\frac{y}{\cos x}\right)'=2x

$ \iff y=(x^2+C)\cos x\quad\text{.for }\exist C\in\Complex

もしくは

$ y'+y\tan x=2x\cos x

$ \iff y'\cos x-y(-\sin x)=2x(\cos x)^2

$ \iff\left(\frac{y}{\cos x}\right)'=(x^2)'

$ \mathrm dy+p'(x)y\mathrm dx=e^{-p(x)}\mathrm d(e^{p(x)}y)

$ (\mathrm D+p'(x))y=e^{-p(x)}\mathrm D(e^{p(x)}y)

これを使えば、

$ y'+p'(x)y=f(x)

$ \iff(e^{p(x)}y)'=e^{p(x)}f(x)

という変形ができる

$ \iff\mathrm d(xy)+3\mathrm d(y^2)=0

$ \iff xy+3y^2=\rm const.

$ \underline{\iff y=\frac{-x\pm\sqrt{x^2-C}}{6}\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ \iff y^4\mathrm d(x^3)+x^3\mathrm d(y^4)-\mathrm dy=0

$ \iff\mathrm d(x^3y^4-y)=0

$ \underline{\iff x^3y^4-y=\rm const.\quad}_\blacksquare

$ y=y'x-(y')^2

$ \iff\begin{dcases}y=y'x-(y')^2\\y'=y''x+y'-2y'y''\end{dcases}

$ \iff\begin{dcases}y=y'x-(y')^2\\y''(x-2y')=0\end{dcases}

$ \iff\exist C\in\Complex:\begin{dcases}y=y'x-(y')^2\\y'=C\lor y'=\frac12x\end{dcases}

$ \iff\exist C\in\Complex:y=C(x-C)\lor y=\frac12 x^2-\frac14x^2=\frac14x^2

$ \underline{\iff y=\frac14x^2\lor\exist C\in\Complex:y=C(x-C)\quad}_\blacksquare

$ y=y'x-\ln y'

$ \iff\begin{dcases}y=y'x-\ln y'\\0=y''\left(x-\frac1{y'}\right)\end{dcases}

$ \iff\exist C\in\Complex:\begin{dcases}y=y'x-\ln y'\\y'=C\lor y'=\frac1x\end{dcases}

$ \underline{\iff\exist C\in\Complex:y=Cx-\ln C\lor y=1+\ln x\quad}_\blacksquare

$ y'+y=xy^3

変数分離はむずい

$ y'=y(xy^2-1)

$ \frac1y\frac1{xy^2-1}\mathrm dy=

$ y'+y=xy^3

$ \iff-\frac12(y^{-2})'+y^{-2}=x\lor y=0

$ \iff-\frac12\left(y^{-2}-x-\frac12\right)'+\left(y^{-2}-x-\frac12\right)=0\lor y=0

$ \iff -\frac12\mathrm d\left(\ln\left|y^{-2}-x-\frac12\right|-2x\right)=0\lor y=0

$ \iff \exist C\in\Complex:y^{-2}-x-\frac12=Ce^{2x}\lor y=0

$ \underline{\iff \exist C\in\Complex:(Ce^{2x}+2x+1)y^2=2\lor y=0\quad}_\blacksquare

$ y'+\frac yx=\frac{x^3}{y^2}

$ \iff(xy)'=\frac{x^4}{y^2}

$ \iff\left(x^3y^3\right)'=3x^6

$ \iff (xy)^3=\frac37 x^7+C\quad\text{.for }\exist C\in\Complex

$ \underline{\iff y^3=\frac37x^4+\frac{C}{x^3}\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

もしくは

$ \iff (y^3)'+3x^{-1}y^3=3x^3

$ \iff (y^3)'+y^3(3\ln|x|)'=3x^3

$ \iff\frac1{|x|^3}(y^3|x|^3)'=3x^3

$ \iff(y^3x^3)'=\frac37(x^7)'

$ \underline{\iff y^3=\frac37x^4+\frac{C}{x^3}\quad\text{.for }\exist C\in\Complex\quad}_\blacksquare

$ y'=(z(\ln x))'=\frac{z'(\ln x)}{x}

$ y''=(y(e^t))''=-\frac{z'(\ln x)}{x^2}+\frac{z''(\ln x)}{x^2}

よって

$ y''+\frac1xy'+\frac4{x^2}y=\left(\frac{2\ln x}{x}\right)^2

$ \iff\frac{z''(\ln x)}{x^2}-\frac{z'(\ln x)}{x^2}+\frac{z'(\ln x)}{x^2}+\frac4{x^2}z(\ln x)=\left(\frac{2\ln x}{x}\right)^2

$ z=z_h+z_p

$ z_h''+4z_h=0\iff z_h=Ae^{2it}+Be^{-2it}

$ z_p''+4z_p=(2t)^2

$ \implies2+4t^2+4bt+4c=4t^2

$ \iff b=0\land c=-\frac12

$ \therefore z_p=t^2-\frac12

$ \therefore z=t^2-\frac12+Ae^{2it}+Be^{-2it}

これと②より

$ \underline{\therefore y=z(\ln x)=(\ln x)^2-\frac12+Ax^{2i}+Bx^{-2i}\quad}_\blacksquare

(7)

$ \iff (D-2)(D+3)y=0

$ \iff (D+3)y=A'e^{2x}\quad\text{.for }\exist A'\in\Complex

$ \iff (e^{3x}y)'=A'e^{2x+3x}\quad\text{.for }\exist A'\in\Complex

$ \iff e^{3x}y=Ae^{2x+3x}+B\quad\text{.for }\exist A,B\in\Complex

$ \underline{\iff y=Ae^{2x}+Be^{-3x}\quad\text{.for }\exist A,B\in\Complex\quad}_\blacksquare

$ \iff (D+3)(D+3)y=0

$ \iff (D+3)y=Ae^{-3x}\quad\text{.for }\exist A\in\Complex

$ \iff (e^{3x}y)'=Ae^{-3x+3x}\quad\text{.for }\exist A\in\Complex

$ =A

$ \iff e^{3x}y=Ax+B\quad\text{.for }\exist A,B\in\Complex

$ \underline{\iff y=Axe^{-3x}+Be^{-3x}\quad\text{.for }\exist A,B\in\Complex\quad}_\blacksquare