[tex]2y`+y=e^x[/tex]
[tex]y`+\frac{1}{2}y=\frac{1}{2}e^x[/tex]
[tex]e^{\frac{1}{2}x} y =\frac{2}{3}e^{\frac{3}{2}x}+C[/tex]
[tex]y =\frac{2}{3}e^{\frac{3}{2}x}\times e^{-\frac{1}{2}x}+C[/tex]
[tex]y(x)= \frac{2}{3}e^x +e^{-\frac{x}{2}}C[/tex]
Fasit sier derimot
[tex]y(x)= \frac{1}{3}e^x +e^{-\frac{x}{2}}C[/tex]
Hvem har rett?
Lett difflikning
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Betelgeuse
- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
Fasiten har nok rett :p
[tex](e^{\frac{1}{2}x}y)\prime = \frac{1}{2}e^{\frac{3}{2}x} \Rightarrow e^{\frac{1}{2}x}y = \frac{1}{2} \int e^{\frac{3}{2}x}dx = \frac{1}{2}\frac{2}{3} e^{\frac{3}{2}x} + C \Rightarrow y = \frac{1}{3}e^x + Ce^{-\frac{1}{2}x}[/tex]
[tex](e^{\frac{1}{2}x}y)\prime = \frac{1}{2}e^{\frac{3}{2}x} \Rightarrow e^{\frac{1}{2}x}y = \frac{1}{2} \int e^{\frac{3}{2}x}dx = \frac{1}{2}\frac{2}{3} e^{\frac{3}{2}x} + C \Rightarrow y = \frac{1}{3}e^x + Ce^{-\frac{1}{2}x}[/tex]
[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]