Uspela sam da rešim prvu, koristeći determinantu i smenu [inlmath]x=u+\alpha[/inlmath], [inlmath]y=v+\beta[/inlmath]
[dispmath]\beta+2=0\quad\Rightarrow\quad\beta=-2\\
\alpha+\beta-1=0\\
\alpha=3\\
y'=\frac{2(v-2+2)^2}{(u+3+v-2-1)^2}=\frac{2v^2}{(u+v)^2}\\
\left.\frac{\mathrm dv}{\mathrm du}=\frac{2v^2}{v^2+2vu+u^2}\quad\right/:u^2\\
\frac{\mathrm dv}{\mathrm du}=\frac{2\frac{v^2}{u^2}}{\frac{v^2}{u^2}+2\frac{v}{u}+1}[/dispmath]
smena: [inlmath]w=\frac{v}{u}[/inlmath]
[dispmath]u\cdot w'+w=\frac{2w^2}{w^2+2w+1}\\
u\frac{\mathrm dw}{\mathrm du}+w=\frac{2w^2}{w^2+2w+1}\\
u\frac{\mathrm dw}{\mathrm du}=\frac{2w^2}{w^2+2w+1}-w\\
u\frac{\mathrm dw}{\mathrm du}=\frac{2w^2-w^3-2w^2-w}{w^2+2w+1}\\
\left.u\frac{\mathrm dw}{\mathrm du}=\frac{-w\left(w^2+1\right)}{(w+1)^2}\quad\right/:\mathrm dw\\
\frac{\mathrm du}{u}=\frac{(w+1)^2\mathrm dw}{-w\left(w^2+1\right)}\\
I=\frac{(w+1)^2}{-w\left(w^2+1\right)}=\frac{A}{-w}+\frac{Bw+C}{w^2+1}=\cdots=\int\frac{1}{-w}\mathrm dw+\int\frac{-2}{w^2+1}\mathrm dw\\
=-\int\frac{1}{w}\mathrm dw-2\int\frac{1}{w^2+1}\mathrm dw\\
=-\ln|w|-2\text{tg}^{-1}w\\
\ln|u|+c=-\ln|w|-2\text{tg}^{-1}w,\;w=\frac{v}{u}=\frac{y+2}{x-3}\\
\ln|uc|=-\ln\left|\frac{v}{u}\right|-2\text{tg}^{-1}\left|\frac{v}{u}\right|\\
\ln|c(y+2)|=-\ln\left|\frac{y+2}{x-3}\right|-2\text{tg}^{-1}\left|\frac{y+2}{x-3}\right|[/dispmath]
a ∧- predstavlja lambdu [inlmath](\lambda)[/inlmath]
Kako da rešim ovu drugu ?
imam [inlmath]y''+4y=2+(8x-1)\sin2x[/inlmath]
[inlmath]\lambda^2+4\lambda=0\quad\Rightarrow\quad\lambda_1=0,\;\lambda_2=-4[/inlmath]
Odatle [inlmath]\left\{1,e^{-4x}\right\}[/inlmath]
[inlmath]b(x)=2+(8x-1)\sin2x[/inlmath] Kako odavde da odredim [inlmath]\alpha[/inlmath] i [inlmath]\beta[/inlmath]?