Respuesta :
Answer:
[tex]y = 10cos (7x) - \frac{4}{7}sin ( 7x )[/tex]
B.
B.
[tex]y = \frac{17}{8}e^4^x - \frac{1}{8}e^-^4^x[/tex]
Step-by-step explanation:
Question 1:
- We are given a homogeneous second order linear ODE as follows:
[tex]y'' + 49y = 0[/tex]
- A pair of independent functions are given as ( y1 ) and ( y2 ):
[tex]y_1 = cos ( 7x )\\\\y_2 = sin ( 7x )[/tex]
- The given ODE is subjected to following initial conditions as follows:
[tex]y ( 0 ) = 10\\\\y ' ( 0 ) = -4[/tex]
- We are to verify that the given independent functions ( y1 ) and ( y2 ) are indeed the solution to the given ODE. If the functions are solutions then find the complete solution of the homogeneous ODE of the form:
[tex]y = c_1y_1 + c_2y_2[/tex]
Solution:-
- To verify the functions are indeed the solution to the given ODE. We will plug the respective derivatives of each function [ y1 and y2 ] into the ODE and prove whether the equality holds true or not.
- Formulate the second derivatives of both functions y1 and y2 as follows:
[tex]y'_1 = -7sin(7x) , y''_1 = -49cos(7x)\\\\y'_2 = -7cos(7x) , y''_2 = -49sin(7x)\[/tex]
- Now plug the second derivatives of each function and the functions itself into the given ODE and verify whether the equality holds true or not.
[tex]y''_1 + 49y_1 = 0\\\\-49cos(7x) + 49cos ( 7x ) = 0\\0 = 0\\\\y''_2 + 49y_2 = 0\\\\-49sin(7x) + 49sin ( 7x ) = 0\\0 = 0\\\\[/tex]
- We see that both functions [ y1 and y2 ] holds true as the solution to the given homogeneous second order linear ODE. Hence, are the solution to given ODE.
- The complete solution to a homogeneous ODE is given in the form as follows:
[tex]y = c_1y_1 + c_2y_2\\\\y = c_1*cos(7x) + c_2*sin(7x)\\[/tex]
- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,
[tex]y (0) = c_1cos ( 0 ) + c_2sin ( 0 ) = 10\\\\y'(0) = -7c_1*sin(0) + 7c_2*cos(0) = -4\\\\c_1 ( 1 ) + c_2 ( 0 ) = 10, c_1 = 10\\\\-7c_1(0) + 7c_2( 1 ) = -4 , c_2 = -\frac{4}{7}[/tex]
- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:
[tex]y = 10cos (7x) - \frac{4}{7}sin ( 7x )[/tex] .... Answer
Question 2
- We are given a homogeneous second order linear ODE as follows:
[tex]y'' -16y =0[/tex]
- A pair of independent functions are given as ( y1 ) and ( y2 ):
[tex]y_1 = e^4^x\\\\y_2 = e^-^4^x[/tex]
- The given ODE is subjected to following initial conditions as follows:
[tex]y( 0 ) = 2\\y'( 0 ) = 9[/tex]
- We are to verify that the given independent functions ( y1 ) and ( y2 ) are indeed the solution to the given ODE. If the functions are solutions then find the complete solution of the homogeneous ODE of the form:
[tex]y = c_1y_1 + c_2y_2[/tex]
Solution:-
- To verify the functions are indeed the solution to the given ODE. We will plug the respective derivatives of each function [ y1 and y2 ] into the ODE and prove whether the equality holds true or not.
- Formulate the second derivatives of both functions y1 and y2 as follows:
[tex]y'_1 = 4e^4^x , y''_1 = 16e^4^x\\\\y'_2 = -4e^-^4^x , y''_2 = 16e^-^4^x[/tex]
- Now substitute the second derivatives of each function and the functions itself into the given ODE and verify whether the equality holds true or not.
[tex]y''_1 - 16y_1 = 0\\\\16e^4^x - 16e^4^x = 0\\\\0 = 0\\\\y''_2 - 16y_2 = 0\\\\16e^-^4^x - 16e^-^4^x = 0\\\\0 = 0[/tex]
- We see that both functions [ y1 and y2 ] holds true as the solution to the given homogeneous second order linear ODE. Hence, are the solution to given ODE.
- The complete solution to a homogeneous ODE is given in the form as follows:
[tex]y = c_1y_1 + c_2y_2\\\\y = c_1*e^4^x + c_2*e^-^4^x[/tex]
- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,
[tex]y ( 0 ) = c_1 * e^0 + c_2 * e^0 = 2\\\\y' ( 0 ) = 4 c_1 * e^0 - 4c_2 * e^0 = 9\\\\c_1 + c_2 = 2 , 4c_1 - 4c_2 = 9\\\\c_1 = \frac{17}{8} , c_2 = -\frac{1}{8}[/tex]
- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:
[tex]y = \frac{17}{8} e^4^x - \frac{1}{8}e^-^4^x[/tex] .... Answer