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Ajutor pls

 
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mihai.salajan



Joined: 17 Jun 2013
Posts: 1

PostPosted: Mon Jun 17, 2013 2:06 pm    Post subject: Ajutor pls Reply with quote

In primul rand doresc sa imi cer scuze adminilor/moderatorilor pentru ca nu sunt convins ca am voie sa creez un topic nou pentru nevoile mele
In alta ordine de idei solicit ajutorul celor pasionati de metematici, urmeaza sa dau un examen din ecuatii diferentiale si sunt putin pe langa subiect profesorul a fost suficient de ingaduitor sa ne prezinte o varianta de lucrare, ceea ce inseamna ca voi primi exercitii similare dar cu mici modificari pe ici pe colo am sa postez o poza cu exercitiile mele daca exista cineva binevoitor si cu inima sa ma ajute si pe mine, v-asi fii foarte recunoscator
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Tue Jan 28, 2014 5:26 pm    Post subject: problema 1 Reply with quote

y(x)=\sin(x)+c, unde c e o constanta reala. Folosim conditia initiala y(0)=1 si obtinem ca c=1.
Prin urmare, y(x)=\sin(x)+1
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Tue Jan 28, 2014 5:49 pm    Post subject: Problema 2 Reply with quote

2) y\prime+3x^2y=e^{x-x^3}. Inmultim relatia cu e^{x^3} si obtinem (e^{x^3}y)\prime=e^x\Rightarrow e^{x^3}y=e^x+c\Rightarrow y=e^{x-x^3}+c e^{-x^3}, unde c este o constanta reala.
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Tue Jan 28, 2014 6:02 pm    Post subject: Reply with quote

3)y\prime=-2xy+y^2e^{x^2+x}. Consider ecuatia asociata \overline{y}\prime=-2x\overline{y}\Rightarrow \overline{y}=ce^{-x^2}(aici c e constanta reala) . Inlocuind in ecuatia Bernoulli(y(x)=c(x)e^{-x^2}, obtinem(dupa calcule) c(x)e^{-x^2}=c(x)^2e^{x-x^2}
Daca c(x)=0\Rightarrow y(x)=0
Daca c\neq 0\Rightarrow c(x)=e^{-x}. Prin urmare, solutiile ecuatiei sunt y_1(x)=0 si y_2(x)=e^{-2x^2}
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Tue Jan 28, 2014 6:14 pm    Post subject: Reply with quote

4)\lambda^5-10\lambda^3+9\lambda=0\Rightarrow \lambda(\lambda-1)(\lambda+1)(\lambda-3)(\lambda+3)=0\Rightarrow \lambda_1=0, \lambda_2=1,\lambda_3=-1, \lambda_4=3,\lambda_5=-3
Solutiile sunt: \phi_1(x)=1,\phi_2(x)=e^x,\phi_3(x)=e^{-x},\phi_4(x)=e^{3x}, \phi_5(x)=e^{-3x}.
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