When à = 0 the differential equation becomes X" = 0 which gives the solution X = ax + b.
Quadratic Equation
When it comes to the concept of polynomial equations, quadratic equations can be said to be a special case. What does solving a quadratic equation mean? We will understand the quadratics and their types once we are familiar with the polynomial equations and their types.
Demand and Supply Function
The concept of demand and supply is important for various factors. One of them is studying and evaluating the condition of an economy within a given period of time. The analysis or evaluation of the demand side factors are important for the suppliers to understand the consumer behavior. The evaluation of supply side factors is important for the consumers in order to understand that what kind of combination of goods or what kind of goods and services he or she should consume in order to maximize his utility and minimize the cost. Therefore, in microeconomics both of these concepts are extremely important in order to have an idea that what exactly is going on in the economy.
I dont understand how X=ax+b. Can you please explain it to me. Thank you
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< Chapter 12.1, Problem 10E >
The given partial differential equation is k = du, k > 0.
Let the solution be u (x, t) = X (x) T (t).
Now, differentiating u partially with respect to x and t gives
u = X"T and du = XT'
Substituting these values in the given partial differential equation
becomes
kX"T = XT'
k = 4
* = 5 =-
where i is the separation constant.
On solving further it becomes,
X" + AX = 0 and T' + AKT = 0
On solving further it gives, T = c2e-kt.
For first differential equation consider three cases:
When 1 = 0 the differential equation becomes X" = 0 which gives
the solution X = ax + b.
Thus, the product solution is,
u (x, t) = X (x) T (t)
= (ax + b) (cze¬dk)
= e-0* (Ax + B)
= Ax + B
where A = ac2 and B = bc2.
When A = -a? < 0, the differential equation becomes X" – aX = 0
which gives the solution X = a cosh ax + b sinh ax.
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