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# Tutor profile: Linus M.

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Linus M.
5 years experience in tutoring field
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## Questions

### Subject:Computer Science (General)

TutorMe
Question:

Perform the following tasks: 1. Write an SQL query to create a a database named Students. 2. Assuming you have more than one database at your disposal, write an sql query to select the above database to be used. 3. Now that you have created a database, create a table named StudentPerformance and inside the table insert random values for Student_Name(should be unique) and Gender. Include student_id (which should be the primary key) NOTE: your table should contain at least five entries(NOT NULL)

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Linus M.

1. CREATE DATABASE Students; 2. USE Students; 3. CREATE TABLE StudentPerformance ( Student_id INT NOT NULL, Student_Name VARCHAR(30) NOT NULL, Gender VARCHAR(30) NOT NULL, PRIMARY KEY (Student_id), UNIQUE (Student_Name) ); 4. INSERT INTO Students (Student_id, Student_Name, Gender) VALUES (1,"Dan", "Male"), (2,"Joan", "Female"), (3,"Mercy", "Female") (4, "Mathew", "Male") (5, "Steven", "Male") (6, "Jane", "Female") (7, "John", "Male");

### Subject:Calculus

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Question:

Solve the following double integral $$\int_1^2\int_0^1(3xy^{2} + 2x^{3}y){d}y{d}x$$

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Linus M.

To begin with, we start by integrating the inner part with respect to y with 1 and 0 as the limits. $$\int_0^1(3xy^{2} + 2x^{3}y){d}y$$ $$=\frac{3xy^{3}}{3} + \frac{2x^{3}y^{2}}{2}\Big|_0^1$$ $$={xy^{3}} + {x^{3}y^{2}}\Big|_0^1$$ Now using the limits $$= (x + x^{3}) - 0$$ We then intergrate with the other limits with respect to x $$=\int_1^2 (x + x^{3}){d}x$$ $$=\frac{x^{2}}{2} + \frac{x^{4}}{2}\Big|_1^2$$ $$= \frac{x^{2} + 4}{2}\Big|_1^2$$ Applying the limits: $$= \frac{4 + 16}{2}- (\frac{1}{2} + \frac{1}{2})$$ $$= 10 - 1$$ $$= 9$$

### Subject:Differential Equations

TutorMe
Question:

Solve the following differential equation $$y' = 2x{e}^{2y}$$ Subject to the initial condition $$y(0) = -1$$

Inactive
Linus M.

Separate variables solving for y on one side and x on the other side $$\frac{dy}{dx} = ( 2x)({e}^{2y})$$ $$\frac{1}{{e}^{2y}}{d}y = ( 2x){d}x$$ Now integrate both sides $$\int\frac{1}{{e}^{2y}}{d}y =\int ( 2x){d}x$$ $$\int {{e}^{-2y}}{d}y =\int ( 2x){d}x$$ $$-\frac{1}{2} {{e}^{-2y}} = {x}^{2} + C$$ Now solve for y $${{e}^{-2y}} =-2( {x}^{2} + C)$$ $${{e}^{-2y}} =-2 {x}^{2} + C$$ $$\ln({{e}^{-2y}}) = \ln(-2 {x}^{2} + C)$$ $$-2y = \ln(-2 {x}^{2} + C)$$ $$y = -\frac{1}{2}\ln(-2x^{2} + C)$$ Now replacing for $$y(0) = -1$$ $$-1 = -\frac{1}{2}\ln(0 + C)$$ $$2 = \ln(0 + C)$$ $$2 = \ln(C)$$ $${{e}^{2}} = {{e}^{\ln(C)}}$$ $${{e}^{2}} = C$$ Since we are solving for y where; $$y = -\frac{1}{2}\ln(-2x^{2} + C)$$ We replace C with $${{e}^{2}}$$ Therefore, final solution is: $$y = -\frac{1}{2}\ln(-2x^{2} + {{e}^{2}})$$

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