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Pirmos eilės tiesinės diferencialinės lygtys - Vikipedija

Pirmos eilės tiesinės diferencialinės lygtys

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Šis straipsnis yra apie Pirmos eilės tiesines diferencialines lygtis.

- $y' + P (x) y = Q (x),$

${dy\over dx}+P(x)y=Q(x),$

y = u v

,

y' = u' v + u v'.

u' v + u v' + P (x) u v = Q (x),

v (u' + P (x) u) + u v' = Q (x);

u' + P (x) u = 0,

${du\over u}=-P(x)dx,$

$u=C_1 e^{-\int P(x)dx};$

$C_1 e^{-\int P(x)dx}v'=Q(x),$

$v'={1\over C_1}Q(x)e^{\int P(x)dx},$

$v=\int {1\over C_1}Q(x)e^{\int P(x)dx}dx+C_2;$

$y=uv=C_1 e^{-\int P(x)dx}(\int {1\over C_1}Q(x)e^{\int P(x)dx}dx+C_2)=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C_1 C_2)=$

$=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C).$

$y'={2y\over x}+x^2 e^x-1,$

$y'-{2\over x}\cdot y=x^2 e^x-1,$

y = u v,

y' = u' v + u v',

$u'v+uv'-{2\over x}uv=x^2 e^x-1,$

$v(u'-{2\over x}u)+uv'=x^2 e^x-1;$

$u'-{2\over x}u=0,$

${du\over u}={2\over x}dx,$

ln | u | = 2ln | x | ,

u = x 2;

x 2 v' = x 2 e x - 1,

$dv=(e^x-{1\over x^2})dx,$

$v=e^x+{1\over x}+C;$

$y=uv=x^2(e^x+{1\over x}+C)=Cx^2+x^2 e^x+x.$

$y' - a y = e b x,$

y = u v,

y' = u' v + u v',

u' v + u v' - a u v = e b x,

v (u' - a u) + u v' = e b x;

u' - a u = 0,

${du\over dx}=au,$

${du\over u}=a\;dx,$

ln | u | = a x,

u = e a x;

e a x v' = e b x,

${dv\over dx}=e^{bx-ax},$

$\int dv=\int e^{(b-a)x}dx,$

$v={1\over b-a}e^{(b-a)x}+C,$ jei $a\neq b$ ir

v = x + C,

jei

a = b,

nes

e 0 = 1;

$y=uv=e^{ax}({e^{(b-a)x}\over b-a}+C)={e^{bx}\over b-a}+Ce^{ax},$ jei $a\neq b$ ir

y = e a x (x + C),

jei

a = b .

- $x y' + P (y) x = Q (y),$

${dx\over dy}+P(y)x=Q(y),$

x = u v,

u = u (y),

v = v (y).

$y'={1\over \cos^2 y-x\tan y},$

${1\over y_x'}=x_y',$

${1\over x_y'}={1\over \cos^2 y-x\tan y},$

x y' = cos 2 y - x tan y,

x y' + x tan y = cos 2 y,

x = x (y),

x = u v,

x y' = u' v + u v' = u'(x) v (x) + u (x) v'(x),

u' v + u v' + u v tan y = cos 2 y,

v (u' + u tan y) + u v' = cos 2 y;

u' + u tan y = 0,

${du\over u}=-\tan y dy,$

ln | u | = ln | cos y | ;

v'cos y = cos 2 y,

v' = cos y,

v = sin y + C;

x = u v = cos y (sin y + C) = sin y cos y + C cos y .

[taisyti] Konstantos variacijos metodas (Lagranžo metodas)

- $y' + P (x) y = Q (x);$

y' + P (x) y = 0,

$y=Ce^{-\int P(x)dx};$

$y=C(x)e^{-\int P(x)dx};$

$y'+P(x)y=C'(x)e^{-\int P(x)dx}+C(x)e^{-\int P(x)dx}\cdot (-P(x))+P(x)C(x)e^{-\int P(x)dx}=Q(x),$

$C'(x)e^{-\int P(x)dx}=Q(x),$

$C'(x)=Q(x)e^{\int P(x)dx},$

$C(x)=\int Q(x)e^{\int P(x)dx}dx+C.$

$y'-{2xy\over 1+x^2}=1+x^2,$ $y | x = 2 = 5;$

$y'-{2xy\over 1+x^2}=0,$

${dy\over dx}={2xy\over 1+x^2},$

$\int {dy\over y}=\int{2x\over 1+x^2}dx,$

$\int {dy\over y}=\int{d(1+x^2)\over 1+x^2},$

ln | y | = ln | 1 + x 2 | + ln | C | ,

y = C (1 + x 2);

y = C (x)(1 + x 2),

$y'=C'(x)\cdot (1+x^2)+C(x)2x;$

$y'-{2xy\over 1+x^2}=C'(x)(1+x^2)+C(x)\cdot 2x-{2xC(x)(1+x^2)\over 1+x^2}=1+x^2,$

$C'(x)\cdot(1+x^2)=1+x^2,$

$C'(x)=1,\; C(x)=x+C;$

y = (x + C)(1 + x 2);

5 = (2 + C)(1 + 2 2),

1 = 2 + C,

C = - 1;

y = (x - 1)(1 + x 2).

${dz\over dx}+{3\over x}z=0,$

$\int{dz\over z}=-3\int{dx\over x},$

ln | z | = - 3ln | x | + ln | C | = ln | C x - 3 | ,

$z={C\over x^3};$

C = C (x),

z = C (x) x - 3,

${dz\over dx}={dC(x)\over dx}{1\over x^3}-{3C(x)\over x^4},$

${dz\over dx}+{3\over x}z={dC(x)\over dx}{1\over x^3}-{3C(x)\over x^4}+{3\over x}{C(x)\over x^3}=x^3,$

${dC(x)\over dx}{1\over x^3}=x^3,$

${dC(x)\over dx}=x^6,$

$\int dC(x)=\int x^6 dx,$

$C(x)={x^7\over 7}+C_1;$

$z={C(x)\over x^3}={{x^7\over 7}+C_1\over x^3}={x^4\over 7}+{C_1\over x^3}.$

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