y''-y=0, y1(t)=e^t , y2(t)=cosht

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 and , 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

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

y" - y

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

So is a solution of the differential equation.

When

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

y" - y

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

So, is a solution of the differential equation.

For First Solution:

is the solution of equation y''-y=0.

For 2nd Solution:

is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution:

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.

First order derivative:

2nd order Derivative:

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence is the solution of equation y''-y=0.

For 2nd Solution:

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.

First order derivative:

2nd order Derivative:

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence is the solution of equation y''-y=0.

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