Respuesta :
Answer:
[tex](a)\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
[tex](b)\ y(t)=\ (1-t)e^{-t}\ -\ 2e[/tex]
Step-by-step explanation:
(a) [tex]y'\ +\ 2y\ =\ te^{-2t},\ y(1)\ =\ 0[/tex]
[tex]=>\ (D+2)y\ =\ te^{-2t}[/tex]
To find the complementary function
D+2 = 0
=> D = -2
So, the complementary function can by given by
[tex]y_c(t)\ =\ C.e^{-2t}[/tex]
Now, to find particular integral
[tex](D+2)y_p(t)\ =\ te^{-2t}[/tex]
[tex]=>y_p(t)\ =\ \dfrac{ te^{-2t}}{D+2}[/tex]
[tex]=\ \dfrac{ te^{-2t}}{-2+2}[/tex]
= not defined
So,
[tex]y_p(t)\ =\ \dfrac{ t^2e^{-2t}}{D^2}[/tex]
[tex]=\ \dfrac{t^2e^{-2t}}{(-2)^2}[/tex]
[tex]=\ \dfrac{t^2e^{-2t}}{4}[/tex]
So, complete solution can be given by
[tex]y(t)\ =\ y_c(t)\ +\ y_p(t)[/tex]
[tex]=> y(t) =\ C.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
As given in question
[tex]=>\ y(1)\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]
[tex]=>\ 0\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex]
[tex]=>\ C\ =\ 4e^2[/tex]
Hence, the complete solution can be give by
[tex]=>\ y(t) =\ 4e^2.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
[tex]=>\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]
(b) [tex]t^{3}y'\ +\ 4t^{2}y\ =\ e^{-t},\ y(-1)\ =\ 0[/tex]
[tex]=>\ y'\ +\ 4t^{-1}y\ =\ t^{-3}e^{-t}[/tex]
Integrating factor can be given by
[tex]I.F\ =\ e^{\int (4t^{-1})dt}[/tex]
[tex]=\ e^{log\ t^4}[/tex]
[tex]=\ t^4[/tex]
Now , the solution of the given differential equation can be given by
[tex]y(t)\times t^4\ =\ \int t^{-3}e^{-t}t^4dt\ +\ C[/tex]
[tex]=>\ y(t)\ =\ \int t.e^{-t}dt\ +\ C[/tex]
[tex]=\ (1-t)e^{-t}\ +\ C[/tex]
According to question
[tex]y(-1)\ =\ (1-(-1))e^1\ +\ C[/tex]
[tex]=>\ 0\ =\ 2e\ +\ C[/tex]
[tex]=>\ C\ =\ -2e[/tex]
Now, the complete solution of the given differential equation cab be given by
[tex]y(t)\ =\ (1-t)e^{-t}\ -\ 2e[/tex]
Answer:
a. [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]
b.[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]
Step-by-step explanation:
We are given that
a.[tex]y'+2y=te^{-2t},y(1)=0[/tex]
Compare with [tex]y'+P(t)y=Q(t)[/tex]
We have P(t)=2,Q(t)=[tex]te^{-2t}[/tex]
Integration factor=[tex]\int e^{2dt}=e^{2t}[/tex]
[tex]y\cdot I.F=\int Q(t)\cdot I.F dt+C[/tex]
Substitute the values then, we get
[tex]y\cdot e^{2t}=\int te^{-2t}\cdot e^{2t} dt+C[/tex]
[tex]y\cdot e^{2t}=\int tdt+C[/tex]
[tex]ye^{2t}=\frac{t^2}{2}+C[/tex]
Substitute the values x=1 and y=0
Then, we get [tex]0\cdot e^2=\frac{1}{2}+C[/tex]
[tex]C=-\frac{1}{2}[/tex]
Substitute the value in the given function
[tex]ye^{2t}=\frac{t^2}{2}-\frac{1}{2}[/tex]
[tex]y=\frac{t^2}{2}e^{-2t}-\frac{1}{2}e^{-2t}[/tex]
Hence, [tex]y(t)=\frac{t^2e^{-2t}}{2}-\frac{1}{2}e^{-2t}[/tex]
b.[tex]t^3y'+4t^2y=e^{-t},y(-1)=0[/tex]
[tex]y'+\frac{4}{t}y=\frac{e^{-t}}{t^3}[/tex]
[tex]P(t)=\frac{4}{t},Q(t)=\frac{e^{-t}}{t^3}[/tex]
I.F=[tex]\int e^{\frac{4}{t}dt}=e^{4lnt}=e^{lnt^4}=t^4[/tex]
[tex]y\cdot \frac{t^4}=\int e^{-t}\frac{t^4}{t^3} dt+C[/tex]
[tex]y\cdot t^4=\int te^{-t}dt+C[/tex]
[tex]yt^4=-te^{-t}+\int e^{-t} dt+C[/tex]
[tex]u\cdot v dt=u\int vdt-\int (\frac{du}{dt}\cdot \int vdt)dt[/tex]
[tex]yt^4=-te^{-t}-e^{-t}+C[/tex]
Substitute the values x=-1,y=0 then, we get
[tex]0=-(-1)e-e+C[/tex]
[tex]C+e-e=0[/tex]
C=0
Substitute the value of C then we get
[tex]yt^4=-te^{-t}-e^{-t}[/tex]
[tex]y=-t^{-3}e^{-t}-t^{-4}e^{-t}[/tex]
