Ordinary First Order Differential Equation: I(x,y)[M(x,y)dx+N(x,y)dy]=0

[jo3ster7]

Hi everyone, i am new here at the forum, i have to register because i have two problems that i can't do, ussually i use symbolab to solve the equations but lately, they are focusing more in a "textual" way to put the problems and that it is really complicated for me, i hope that you can lend me a hand with this, i would like to know how to declare it on software like symbolab or wolfram mathematica to get the right answer.

this is one of the problems:

This type of questions consists of a statement, problem or context from which four options numbered from 1 to 4 are proposed, you must select the combination of two options that adequately answers the question and mark it on the answer sheet, according to with the following information:
• Mark A if 1 and 2 are correct.
• Mark B if 1 and 3 are correct.
• Mark C if 2 and 4 are correct.
• Mark D if 3 and 4 are correct.

A. One of the methods of solving ordinary differential equations of the first order is the integration factor, which involves a factor , that by multiplying the O.D.E. the transformation allowed by direct integration or transformation into an O.D.E. Exact That is, if it is true that is exact.

A procedure for supplying a medicament in the blood, that is, permanently by an injection technique. This method can be modeled by the differential equation , where C(t)= Drug concentration at each instant t, plus P, J and k are constants that represent the characteristics of the process and patient-specific conditions.

Given the above information, it can be stated that:
When solving the model that meets the initial condition C(0)=1, we obtain:
The general solution, which is:
The particular solution, which is:
The general solution, which is:
The particular solution, which is:


I Will post the another problema later, because i can not attach more images. I really appreciate the help, i am really bad with maths,

Following the given advice, i put the other excersice in plain text:

b. A differential expression of the form M (x, y) dx + N (x, y) of is known as the exact differential equation in a region R of the xy plane, if it corresponds corresponds to the differential function of function f (x, y) Defined in R. A first order differential equation of the form: M (x, y) dx + N (x, y) dy = 0 It is said to be an exact equation if the expression on the left side is an exact exact one. The criterion for an exact verification is: Let M (x, y) and N (x, y) be continuous and have first continuous partial derivatives in a rectangular region R defined by a <x <b, c <and <d. So, a necessary and sufficient condition for M (x, y) dx + N (x, y) dy to be an exact verification is:
/M / ∂y = ∂N / ∂x
According to the previous information to the problem the differential equation (5x + 4y) dx + (4x-8y ^ 3) dy = 0 by this method we obtain that the values ​​for ∂M / ∂y, ∂N / ∂x and the solution The general differential equation are respectively:


∂M / ∂y = ∂N / ∂x = 5 + 24y ^ 2
f (x, y) = (5/2) x ^ 2 + 5xy + 24y ^ 2
f (x, y) = (5/2) x ^ 2 + 4xy-2y ^ 4
∂M / ∂y = ∂N / ∂x = 4

Thank you in advance,

John.

 

[mmm4444bot]

… Will post the another problema later, because i can not attach more images …

Hi. You don't need to post images of short expressions like I(x,y).

:idea: If the image limit is a problem, you could type out other expressions, too.

I(x,y) [M(x,y) dx + N(x,y) dy] = 0

P dC - (J - k*C(t)) dt = 0

P*e^(k*t/P)*C - (JP/k) e^(kt/P) = α

C(t) = 1/e^(kt/P) (1 - J/k + J/k e^(kt/P))

P*e^(-kt)*C + (JP/k) e^(-kt) = α

C(t) = J/k (1 - e^(-kt/P))

 

[jo3ster7]

Hi. You don't need to post images of short expressions like I(x,y).

:idea: If the image limit is a problem, you could type out other expressions, too.

I(x,y) [M(x,y) dx + N(x,y) dy] = 0

P dC - (J - k*C(t)) dt = 0

P*e^(k*t/P)*C - (JP/k) e^(kt/P) = α

C(t) = 1/e^(kt/P) (1 - J/k + J/k e^(kt/P))

P*e^(-kt)*C + (JP/k) e^(-kt) = α

C(t) = J/k (1 - e^(-kt/P))

Thank you, i did that for the second problem