Answer:
a.[tex]w(t)=-12e^{2t}[/tex]
b.[tex] y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}[/tex]
Step-by-step explanation:
We have a differential equation
y''-2 y'-35 y=0
Auxillary equation
[tex](D^2-2D-35)=0[/tex]
By factorization method we are finding the solution
[tex]D^2-7D+5D-35=0[/tex]
[tex](D-7)(D+5)=0[/tex]
Substitute each factor equal to zero
D-7=0 and D+5=0
D=7 and D=-5
Therefore ,
General solution is
[tex]y(x)=C_1e^{7t}+C_2e^{-5t}[/tex]
Let [tex]y_1=e^{7t} \;and \;y_2=e^{-5t}[/tex]
We have to find Wronskian
[tex]w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}[/tex]
Substitute values then we get
[tex]w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}[/tex]
[tex]w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}[/tex]
[tex]w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}[/tex]
a.[tex]w(t)=-12e^{2t}[/tex]
We are given that y(0)=-7 and y'(0)=23
Substitute the value in general solution the we get
[tex]y(0)=C_1+C_2[/tex]
[tex]C_1+C_2=-7[/tex]....(equation I)
[tex]y'(t)=7C_1e^{7t}-5C_2e^{-5t}[/tex]
[tex]y'(0)=7C_1-5C_2[/tex]
[tex]7C_1-5C_2=23[/tex]......(equation II)
Equation I is multiply by 5 then we subtract equation II from equation I
Using elimination method we eliminate[tex] C_1[/tex]
Then we get [tex]C_2=-\frac{5}{2}[/tex]
Substitute the value of [tex] C_2 [/tex] in I equation then we get
[tex] C_1-\frac{5}{2}=-7[/tex]
[tex] C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}[/tex]
Hence, the general solution is
b.[tex] y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}[/tex]
