In each of problems 5 through 10, verify that each given function is a solution of the differential equation.
y''-y=0, y1(t)=e^t , y2(t)=cosht

Answers

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Determine which function is a solution to the differential equation xy ' + 3y = 0. none of these

x⁻²

Step-by-step explanation:

If y = e²ˣ, y' = 2e²ˣ.

2xe²ˣ + 2e²ˣ ≠ 0

If y = x², y' = 2x.

2x² + 2x² ≠ 0

If y = x⁻², y' = -2x⁻³.

-2x⁻² + 2x⁻² = 0

Since L.H.S = R.H.S = 0, for both y_{1} (t) = e^{t} and y_{2} (t) = cosh(t), y₁ and y₂ both satisfy the equation y" - y = 0 and are thus solutions to the differential equation.

Step-by-step explanation:

To check whether the given functions are solutions the given differential equation, we differentiate the functions and then insert it into the given equation.

So y" - y = 0 and  

y_{1} (t) = e^{t}\\y_{1}' (t) = e^{t}\\ y_{1}" (t) = e^{t}

Substituting these values of y and y" into the left hand side of the equation, we have

y" - y  

y_{1}" (t) - y_{1} (t) = e^{t} - e^{t} = 0

Since L.H.S = R.H.S

So y_{1} (t) = e^{t} is a solution of the differential equation.

When

y_{2} (t) = cosh(t)\\ y_{2}'(t) = sinh(t) \\y_{2}"(t) = cosh(t)

Substituting y and y" into the left hand side of the equation, we have

y" - y

y_{2}"(t) - y_{2} (t) = cosh(t) - cosh(t) = 0

Since L.H.S = R.H.S

So, y_{2} (t) = cosh(t) is a solution of the differential equation.

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.



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