Matematika/Pirmos eilės tiesinės diferencialinės lygtys

Šis straipsnis yra apie Pirmos eilės tiesines diferencialines lygtis.

• $y^{\prime }+P(x)y=Q(x),$

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

$y=uv$, $y^{\prime }=u^{\prime }v+u{v}^{\prime }.$

$u^{\prime }v+u{v}^{\prime }+P(x)uv=Q(x),$

$v(u^{\prime }+P(x)u)+u{v}^{\prime }=Q(x);$

$u^{\prime }+P(x)u=0,$

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

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

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

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

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

$y=uv=C_1{e}^{-\int P(x)dx}(\int \frac{1}{C_1}Q(x)e^{\int P(x)dx}dx+C_2)=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C_1{C}_2)=$

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

• $y^{\prime }=\frac{2y}{x}+x^2{e}^x-1,$

$y^{\prime }-\frac{2}{x}\cdot y=x^2{e}^x-1,$

$y=uv,$ $y^{\prime }=u^{\prime }v+u{v}^{\prime },$

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

$v(u^{\prime }-\frac{2}{x}u)+u{v}^{\prime }=x^2{e}^x-1;$

$u^{\prime }-\frac{2}{x}u=0,$

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

$\ln |u|=2\ln |x|,$

$u=x^2;$

$x^2{v}^{\prime }=x^2{e}^x-1,$

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

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

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

• $y^{\prime }-ay=e^{bx},$

$y=uv,$ $y^{\prime }=u^{\prime }v+u{v}^{\prime },$

$u^{\prime }v+u{v}^{\prime }-auv=e^{bx},$

$v(u^{\prime }-au)+u{v}^{\prime }=e^{bx};$

$u^{\prime }-au=0,$

$\frac{du}{dx}=au,$

$\frac{du}{u}=adx,$

$\ln |u|=ax,$

$u=e^{ax};$

$e^{ax}{v}^{\prime }=e^{bx},$

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

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

$v=\frac{1}{b-a}{e}^{(b-a)x}+C,$ jei $a\ne b$ ir $v=x+C,$ jei $a=b,$ nes $e^0=1;$

$y=uv=e^{ax}(\frac{e^{(b-a)x}}{b-a}+C)=\frac{e^{bx}}{b-a}+C{e}^{ax},$ jei $a\ne b$ ir $y=e^{ax}(x+C),$ jei $a=b.$

• ${x_y}^{\prime }+P(y)x=Q(y),$

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

$x=uv,$ $u=u(y),$ $v=v(y).$

• $y^{\prime }=\frac{1}{{\cos }^{2}y-x\tan y},$

$\frac{1}{{y_x}^{\prime }}={x_y}^{\prime },$

$\frac{1}{{x_y}^{\prime }}=\frac{1}{{\cos }^{2}y-x\tan y},$

${x_y}^{\prime }={\cos }^{2}y-x\tan y,$

${x_y}^{\prime }+x\tan y={\cos }^{2}y,$

$x=x(y),$ $x=uv,$ ${x_y}^{\prime }=u^{\prime }v+u{v}^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x),$

$u^{\prime }v+u{v}^{\prime }+uv\tan y={\cos }^{2}y,$

$v(u^{\prime }+u\tan y)+u{v}^{\prime }={\cos }^{2}y;$

$u^{\prime }+u\tan y=0,$

$\frac{du}{u}=-\tan ydy,$

$\ln |u|=\ln |\cos y|;$

$v^{\prime }\cos y={\cos }^{2}y,$

$v^{\prime }=\cos y,$

$v=\sin y+C;$

$x=uv=\cos y(\sin y+C)=\sin y\cos y+C\cos y.$

Konstantos variacijos metodas (Lagranžo metodas)

• $y^{\prime }+P(x)y=Q(x);$

$y^{\prime }+P(x)y=0,$

$\frac{y^{\prime }}{y}=-P(x),$

$\ln y+\ln C=-\int P(x)dx,$

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

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

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

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

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

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

• $y^{\prime }-\frac{2xy}{1+x^2}=1+x^2,$ $y{|}_{x=2}=5;$

$y^{\prime }-\frac{2xy}{1+x^2}=0,$

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

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

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

$\ln |y|=\ln |1+x^2|+\ln |C|,$

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

$y=C(x)(1+x^2),$ $y^{\prime }=C^{\prime }(x)\cdot (1+x^2)+C(x)2x;$

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

$C^{\prime }(x)\cdot (1+x^2)=1+x^2,$

$C^{\prime }(x)=1,C(x)=x+C;$

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

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

$1=2+C,$

$C=-1;$

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

• $\frac{dz}{dx}+\frac{3}{x}z=0,$

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

$\ln |z|=-3\ln |x|+\ln |C|=\ln |C{x}^{-3}|,$

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

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

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

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

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

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

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

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