Tavallinen differentiaaliyhtälö

Tavallinen differentiaaliyhtälö on differentiaaliyhtälö, jossa on ainoastaan yksi muuttuja.

Määritelmä

Olkoon F muuttujan x funktio ja olkoon y=y(x) yhden muuttujan funktio. Tällöin yhtälöä

F ( x , y ( x ) , y ( x ) , , y ( n 1 ) ( x ) ) = y ( n ) ( x ) {\displaystyle F\left(x,y(x),y'(x),\ldots ,y^{(n-1)}(x)\right)=y^{(n)}(x)}

kutsutaan tavalliseksi differentiaaliyhtälöksi, jonka kertaluku on n.[1]

Eräiden yhtälöiden ratkaisuja

Separoituvat yhtälöt

Yhtälö Ratkaisutapa Ratkaisu
Ensimmäinen kertaluku, x ja y separoituvia[2]

P 1 ( x ) Q 1 ( y ) + P 2 ( x ) Q 2 ( y ) d y d x = 0 P 1 ( x ) Q 1 ( y ) d x + P 2 ( x ) Q 2 ( y ) d y = 0 {\displaystyle {\begin{aligned}P_{1}(x)Q_{1}(y)+P_{2}(x)Q_{2}(y)\,{\frac {dy}{dx}}&=0\\P_{1}(x)Q_{1}(y)\,dx+P_{2}(x)Q_{2}(y)\,dy&=0\end{aligned}}}

Separointi (jakaminen tulolla P2Q1). x P 1 ( λ ) P 2 ( λ ) d λ + y Q 2 ( λ ) Q 1 ( λ ) d λ = C {\displaystyle \int ^{x}{\frac {P_{1}(\lambda )}{P_{2}(\lambda )}}\,d\lambda +\int ^{y}{\frac {Q_{2}(\lambda )}{Q_{1}(\lambda )}}\,d\lambda =C}
Ensimmäinen kertaluku, x separoituva[3]

d y d x = F ( x ) d y = F ( x ) d x {\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=F(x)\\dy&=F(x)\,dx\end{aligned}}}

Integrointi. y = x F ( λ ) d λ + C {\displaystyle y=\int ^{x}F(\lambda )\,d\lambda +C}
Ensimmäinen kertaluku, y separoituva[3]

d y d x = F ( y ) d y = F ( y ) d x {\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=F(y)\\dy&=F(y)\,dx\end{aligned}}}

Separointi (jakaminen F:llä). x = y d λ F ( λ ) + C {\displaystyle x=\int ^{y}{\frac {d\lambda }{F(\lambda )}}+C}
Ensimmäinen kertaluku, x ja y separoituvia[3]

P ( y ) d y d x + Q ( x ) = 0 P ( y ) d y + Q ( x ) d x = 0 {\displaystyle {\begin{aligned}P(y){\frac {dy}{dx}}+Q(x)&=0\\P(y)\,dy+Q(x)\,dx&=0\end{aligned}}}

Integrointi. y P ( λ ) d λ + x Q ( λ ) d λ = C {\displaystyle \int ^{y}P(\lambda )\,d\lambda +\int ^{x}Q(\lambda )\,d\lambda =C}

Ensimmäisen kertaluvun yhtälöt

Yhtälö Ratkaisutapa Ratkaisu
Ensimmäinen kertaluku, homogeeninen[3]

d y d x = F ( y x ) {\displaystyle {\frac {dy}{dx}}=F\left({\frac {y}{x}}\right)}

Sijoita y = ux ja separoi u ja x. ln ( C x ) = y / x d λ F ( λ ) λ {\displaystyle \ln(Cx)=\int ^{y/x}{\frac {d\lambda }{F(\lambda )-\lambda }}}
Ensimmäinen kertaluku, separoituva[2]

y M ( x y ) + x N ( x y ) d y d x = 0 y M ( x y ) d x + x N ( x y ) d y = 0 {\displaystyle {\begin{aligned}yM(xy)+xN(xy)\,{\frac {dy}{dx}}&=0\\yM(xy)\,dx+xN(xy)\,dy&=0\end{aligned}}}

Separointi (jakaminen xy:llä).

ln ( C x ) = x y N ( λ ) d λ λ [ N ( λ ) M ( λ ) ] {\displaystyle \ln(Cx)=\int ^{xy}{\frac {N(\lambda )\,d\lambda }{\lambda [N(\lambda )-M(\lambda )]}}}

Jos N = M, ratkaisu on xy = C.

Eksakti, ensimmäinen kertaluku[3]

M ( x , y ) d y d x + N ( x , y ) = 0 M ( x , y ) d y + N ( x , y ) d x = 0 {\displaystyle {\begin{aligned}M(x,y){\frac {dy}{dx}}+N(x,y)&=0\\M(x,y)\,dy+N(x,y)\,dx&=0\end{aligned}}}

jossa M y = N x {\displaystyle {\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}}

Integrointi. F ( x , y ) = x M ( λ , y ) d λ + y Y ( λ ) d λ = y N ( x , λ ) d λ + x X ( λ ) d λ = C {\displaystyle {\begin{aligned}F(x,y)&=\int ^{x}M(\lambda ,y)\,d\lambda +\int ^{y}Y(\lambda )\,d\lambda \\&=\int ^{y}N(x,\lambda )\,d\lambda +\int ^{x}X(\lambda )\,d\lambda =C\end{aligned}}}

jossa

Y ( y ) = N ( x , y ) y x M ( λ , y ) d λ {\displaystyle Y(y)=N(x,y)-{\frac {\partial }{\partial y}}\int ^{x}M(\lambda ,y)\,d\lambda }
ja
X ( x ) = M ( x , y ) x y N ( x , λ ) d λ {\displaystyle X(x)=M(x,y)-{\frac {\partial }{\partial x}}\int ^{y}N(x,\lambda )\,d\lambda }

Epäeksakti, ensimmäinen kertaluku[3]

M ( x , y ) d y d x + N ( x , y ) = 0 M ( x , y ) d y + N ( x , y ) d x = 0 {\displaystyle {\begin{aligned}M(x,y){\frac {dy}{dx}}+N(x,y)&=0\\M(x,y)\,dy+N(x,y)\,dx&=0\end{aligned}}}

jossa M x N y {\displaystyle {\frac {\partial M}{\partial x}}\neq {\frac {\partial N}{\partial y}}}

Kerroin μ(x, y), jolle

( μ M ) y = ( μ N ) x {\displaystyle {\frac {\partial (\mu M)}{\partial y}}={\frac {\partial (\mu N)}{\partial x}}}

Sopivalle μ(x, y)

F ( x , y ) = x μ ( λ , y ) M ( λ , y ) d λ + y Y ( λ ) d λ = y μ ( x , λ ) N ( x , λ ) d λ + x X ( λ ) d λ = C {\displaystyle {\begin{aligned}F(x,y)=&\int ^{x}\mu (\lambda ,y)M(\lambda ,y)\,d\lambda +\int ^{y}Y(\lambda )\,d\lambda \\=&\int ^{y}\mu (x,\lambda )N(x,\lambda )\,d\lambda +\int ^{x}X(\lambda )\,d\lambda =C\end{aligned}}}

jossa

Y ( y ) = N ( x , y ) y x μ ( λ , y ) M ( λ , y ) d λ {\displaystyle Y(y)=N(x,y)-{\frac {\partial }{\partial y}}\int ^{x}\mu (\lambda ,y)M(\lambda ,y)\,d\lambda }
ja
X ( x ) = M ( x , y ) x y μ ( x , λ ) N ( x , λ ) d λ {\displaystyle X(x)=M(x,y)-{\frac {\partial }{\partial x}}\int ^{y}\mu (x,\lambda )N(x,\lambda )\,d\lambda }

Toisen kertaluvun yhtälöt

Yhtälö Ratkaisutapa Ratkaisu
Toinen kertaluku[4]

d 2 y d x 2 = F ( y ) {\displaystyle {\frac {d^{2}y}{dx^{2}}}=F(y)}

Kerro 2 d y d x {\displaystyle 2{\frac {dy}{dx}}} :llä, sijoita 2 d y d x d 2 y d x 2 = d d x ( d y d x ) 2 {\displaystyle 2{\frac {dy}{dx}}{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)^{2}} ja integroi kahdesti. x = ± y d λ 2 λ F ( ε ) d ε + C 1 + C 2 {\displaystyle x=\pm \int ^{y}{\frac {d\lambda }{\sqrt {2\int ^{\lambda }F(\varepsilon )\,d\varepsilon +C_{1}}}}+C_{2}}

Lineaariset n:nnen kertaluvun yhtälöt

Yhtälö Ratkaisutapa Ratkaisu
Ensimmäinen kertaluku, lineaarinen, epähomogeeninen, kertoimet:[3]

d y d x + P ( x ) y = Q ( x ) {\displaystyle {\frac {dy}{dx}}+P(x)y=Q(x)}

Kerroin: e x P ( λ ) d λ . {\displaystyle e^{\int ^{x}P(\lambda )\,d\lambda }.}

y = e x P ( λ ) d λ [ x e λ P ( ε ) d ε Q ( λ ) d λ + C ] {\displaystyle y=e^{-\int ^{x}P(\lambda )\,d\lambda }\left[\int ^{x}e^{\int ^{\lambda }P(\varepsilon )\,d\varepsilon }Q(\lambda )\,d\lambda +C\right]}

Toinen kertaluku, lineaarinen, epähomogeeninen, kertoimet:

d 2 y d x 2 + 2 p ( x ) d y d x + ( p ( x ) 2 + p ( x ) ) y = q ( x ) {\displaystyle {\frac {d^{2}y}{dx^{2}}}+2p(x){\frac {dy}{dx}}+\left(p(x)^{2}+p'(x)\right)y=q(x)}

Kerroin: e x P ( λ ) d λ {\displaystyle e^{\int ^{x}P(\lambda )\,d\lambda }} y = e x P ( λ ) d λ [ x ( ξ e λ P ( ε ) d ε Q ( λ ) d λ ) d ξ + C 1 x + C 2 ] {\displaystyle y=e^{-\int ^{x}P(\lambda )\,d\lambda }\left[\int ^{x}\left(\int ^{\xi }e^{\int ^{\lambda }P(\varepsilon )\,d\varepsilon }Q(\lambda )\,d\lambda \right)d\xi +C_{1}x+C_{2}\right]}
Toinen kertaluku, lineaarinen, epähomogeeninen, vakiokertoimet[5]

d 2 y d x 2 + b d y d x + c y = r ( x ) {\displaystyle {\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=r(x)}

y = y c + y p {\displaystyle y=y_{c}+y_{p}}

Jos b 2 > 4 c {\displaystyle b_{2}>4c} :

y c = C 1 e x 2 ( b + b 2 4 c ) + C 2 e x 2 ( b b 2 4 c ) {\displaystyle y_{c}=C_{1}e^{-{\frac {x}{2}}\,\left(b+{\sqrt {b^{2}-4c}}\right)}+C_{2}e^{-{\frac {x}{2}}\,\left(b-{\sqrt {b^{2}-4c}}\right)}}

Jos b 2 = 4 c {\displaystyle b_{2}=4c} :

y c = ( C 1 x + C 2 ) e b x 2 {\displaystyle y_{c}=(C_{1}x+C_{2})e^{-{\frac {bx}{2}}}}

Jos b 2 < 4 c {\displaystyle b_{2}<4c} :

y c = e b x 2 [ C 1 sin ( x 4 c b 2 2 ) + C 2 cos ( x 4 c b 2 2 ) ] {\displaystyle y_{c}=e^{-{\frac {bx}{2}}}\left[C_{1}\sin \left(x\,{\frac {\sqrt {4c-b^{2}}}{2}}\right)+C_{2}\cos \left(x\,{\frac {\sqrt {4c-b^{2}}}{2}}\right)\right]}

Kertaluku n, lineaarinen, epähomogeeninen, vakiokertoimet[5]

j = 0 n b j d j y d x j = r ( x ) {\displaystyle \sum _{j=0}^{n}b_{j}{\frac {d^{j}y}{dx^{j}}}=r(x)}

y = y c + y p {\displaystyle y=y_{c}+y_{p}}

y c = j = 1 n ( = 1 k j C j , x 1 ) e α j x {\displaystyle y_{c}=\sum _{j=1}^{n}\left(\sum _{\ell =1}^{k_{j}}C_{j,\ell }x^{\ell -1}\right)e^{\alpha _{j}x}}
jossa
C j e α j x = C j e χ j x cos ( γ j x + φ j ) {\displaystyle C_{j}e^{\alpha _{j}x}=C_{j}e^{\chi _{j}x}\cos(\gamma _{j}x+\varphi _{j})}


Lähteet

  1. Gyllenberg, Mats; Lamberg, Lasse; Ola, Petri; Piiroinen, Petteri; Häsä, Jokke: Tavalliset differentiaaliyhtälöt. Matematiikan ja tilastotieteen laitos, Helsingin yliopisto, 2016.
  2. a b Mathematical Handbook of Formulas and Tables (3rd edition), S. Lipschutz, M. R. Spiegel, J. Liu, Schaum's Outline Series, 2009, ISC_2N 978-0-07-154855-7
  3. a b c d e f g Elementary Differential Equations and Boundary Value Problems (4th Edition), W.E. Boyce, R.C. Diprima, Wiley International, John Wiley & Sons, 1986, ISBN: 0-471-83824-1
  4. Further Elementary Analysis, R. Porter, G.Bell & Sons (London), 1978, ISBN: 0-7135-1594-5
  5. a b Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISC_2N 978-0-521-86153-3