Question1:

Given general solution is: , and
Substituting the above in the linear differential homogeneous equation, we have

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Since , is the solution of linear differential homogeneous equation.

Question 2:
, and
This equation is linear homogeneous equations with constant coefficients. It is linear because no term has one of y or y’ or y” raised to any power except the first power and is homogeneous because every nonzero term contains y or y’ or y”. This expression is multiplied by constants, hence the “constant coefficients” in the name.
A solution of the above differential equation is:
………. (2)
On substituting in the above differential equation, we have
Or
Since cannot be zero,

The general solution is:

Given and , thus
and
From above two equations, we have
and
Thus, the solution is:

Question 3:
, ,
This equation is linear homogeneous equations with constant coefficients. It is linear because no term has one of y or y’ or y” raised to any power except the first power and is homogeneous because every nonzero term contains y or y’ or y”. This expression is multiplied by constants, hence the “constant coefficients” in the name.
A solution of the above differential equation is:
………. (2)
On substituting in the above differential equation, we have
Or
Since cannot be zero,

The general solution is:

Given that ,
and
From above two equations, we have
and
Thus, the general solution is:

Question 4:

This is in the form of second order Euler Cauchy equation
Let ; ;
Substituting above values into original equation we get

Auxiliary equation is:

Solution of given equation is:

Question 5:

; ,
Let us suppose solution to this Euler Cauchy equation is:

Substituting in the above differential equation, we have

Since we assume ; the polynomial is zero.

The general solution of the above equation is:

As given, ,
and

From above two equations, we have
And
Thus, the general solution of the differential equation is: 