Mathematics, 01.07.2020 15:01 182075

A homogeneous second-order linear differential equation, two functions y 1y1 and y 2y2, and a pair of initial conditions are given. First verify that y 1y1 and y 2y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial conditions. Primes denote derivatives with respect to x. y'' + 49y = 0; y1 = cos(7x) y2 = sin(7x); y(0) = 10 y(0)=-4
1.Why is the function y, = e * a solution to the differential equation?
A. The function y1 =e 4X is a solution because when the function and its indefinite integral, , are substituted into the equation, the result is a true statement.
B. The function y1 = e 4X is a solution because when the function and its second derivative, y1" = 16 e 4x, are substituted into the equation, the result is a true statement.
2. Why is the function y2 solution the differential equation?
A. The function y2 = e 4x is a solution because when the function and its indefinite integral, are substituted into the equation, the result a true statement. The function y2 = e 4X is a solution because when the function and its second derivative, y2" = 16 e -4x are substituted into the equation, the result is a true statement. The particular solution of the form y = c, y, +c, y2 that satisfies the initial conditions y(0) 2 and y'(0) = 9 is y =.

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Answer from: ayeheavymetal

$y = 10cos (7x) - \frac{4}{7}sin ( 7x )$

$y = \frac{17}{8}e^4^x - \frac{1}{8}e^-^4^x$

Step-by-step explanation:

Question 1:

- We are given a homogeneous second order linear ODE as follows:

y'' + 49y = 0

- A pair of independent functions are given as ( y1 ) and ( y2 ):

$y_1 = cos ( 7x )\\\\y_2 = sin ( 7x )$

- The given ODE is subjected to following initial conditions as follows:

$y ( 0 ) = 10\\\\y ' ( 0 ) = -4$

- 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:

y = c_1y_1 + c_2y_2

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:

$y'_1 = -7sin(7x) , y''_1 = -49cos(7x)\\\\y'_2 = -7cos(7x) , y''_2 = -49sin(7x)\$

- 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.

$y''_1 + 49y_1 = 0\\\\-49cos(7x) + 49cos ( 7x ) = 0\\0 = 0\\\\y''_2 + 49y_2 = 0\\\\-49sin(7x) + 49sin ( 7x ) = 0\\0 = 0\\\\$

- 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:

$y = c_1y_1 + c_2y_2\\\\y = c_1*cos(7x) + c_2*sin(7x)\\$

- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,

$y (0) = c_1cos ( 0 ) + c_2sin ( 0 ) = 10\\\\y'(0) = -7c_1*sin(0) + 7c_2*cos(0) = -4\\\\c_1 ( 1 ) + c_2 ( 0 ) = 10, c_1 = 10\\\\-7c_1(0) + 7c_2( 1 ) = -4 , c_2 = -\frac{4}{7}$

- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:

$y = 10cos (7x) - \frac{4}{7}sin ( 7x )$ .... Answer

Question 2

- We are given a homogeneous second order linear ODE as follows:

y'' -16y =0

- A pair of independent functions are given as ( y1 ) and ( y2 ):

$y_1 = e^4^x\\\\y_2 = e^-^4^x$

- The given ODE is subjected to following initial conditions as follows:

$y( 0 ) = 2\\y'( 0 ) = 9$

y = c_1y_1 + c_2y_2

Solution:-

- Formulate the second derivatives of both functions y1 and y2 as follows:

$y'_1 = 4e^4^x , y''_1 = 16e^4^x\\\\y'_2 = -4e^-^4^x , y''_2 = 16e^-^4^x$

- 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.

$y''_1 - 16y_1 = 0\\\\16e^4^x - 16e^4^x = 0\\\\0 = 0\\\\y''_2 - 16y_2 = 0\\\\16e^-^4^x - 16e^-^4^x = 0\\\\0 = 0$

- 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:

$y = c_1y_1 + c_2y_2\\\\y = c_1*e^4^x + c_2*e^-^4^x$

- To complete the above solution we need to determine the constants [ c1 and c2 ] using the initial conditions given. Therefore,

$y ( 0 ) = c_1 * e^0 + c_2 * e^0 = 2\\\\y' ( 0 ) = 4 c_1 * e^0 - 4c_2 * e^0 = 9\\\\c_1 + c_2 = 2 , 4c_1 - 4c_2 = 9\\\\c_1 = \frac{17}{8} , c_2 = -\frac{1}{8}$

- Now we can write the complete solution to the given homogeneous second order linear ODE as follows:

$y = \frac{17}{8} e^4^x - \frac{1}{8}e^-^4^x$ .... Answer

Answer from: Quest

answer: b i think

step-by-step explanation:

Answer from: Quest

the answer is two. hope this !

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A homogeneous second-order linear differential equation, two functions y 1y1 and y 2y2, and a pair o...