You will need

- - notebook;
- - handle

Instruction

1

If the equation is written in the form: dy/dx = q(x)/n(y) belong to the category of differential equations with separable variables. They can be solved by writing the condition in the differentials according to the following scheme: n(y)dy = q(x)dx. Then integrate both parts. In some cases the decision is recorded in the form of integrals, taken from known functions. For example, if dy/dx = x/y, you get q(x) = x, n(y) = y. Write it in the form ydy = xdx and integrate. You should get y^2 = x^2 + c.

2

Linear

**equations**consider the equation "first degree." Unknown function with its derivatives included in such equation only in the first degree. Linear differential equation has the form dy/dx + f(x) = j(x), where f(x) and g(x) is a function depending on x. The solution is written with integrals, is taken from the well-known functions.3

Note that a differential equation is equation of second order (contains second derivatives) Thus, for example, is the equation of simple harmonic motion written in the form of the General formula: md 2x/dt 2 = –kx. Such equations have, in General, private decisions. The equation of simple harmonic motion is an example of the important class of linear differential equations with constant coefficient.

4

Let's consider a more common example (second order) equations, where y and z are predetermined constant, f(x) is a given function. Such equations can be solved in different ways, for example, by means of integral transformations. The same can be said about linear equations of higher order with constant coefficients.

5

Please note that the equations that contain the unknown functions and their derivatives standing in a degree above the first, called non-linear. The solution of nonlinear equations is quite complex and therefore, is used for each your particular case.

# Advice 2: How to determine the form of the differential equation

To determine the differential equations required in order to choose the appropriate each case the method of solution. Classification of species is quite large, and the decision is based on the methods of integration.

Instruction

1

The necessity of differential equations arises when the well-known properties of the function, and she remains an unknown quantity. Often this situation occurs in the study of physical processes. Function properties are described by its derivative, or differential, so the only way the problem is integration. Before proceeding to the solution, we need to determine the form of the differential equation.

2

There are several types of differential equations, the simplest of them is the expression y’ = f(x), where y’ = dy/DX. In addition, this type can be given equality f(x)•y’ = g(x), i.e. y’ = g(x)/f(x). Of course, this is only possible provided that f(x) becomes zero. Example: 3^x•y’ = x2 – 1 → y’ = (x2 - 1)/3^H.

3

Differential equation with separated variables are so named because the derivative of y’ in this case literally split in two components dy and DX, which are on opposite sides of the equal sign. This equation of the form f(u)•dy = g(x)•DX. Example: (Y2 – sin y)•dy = tg x/(x - 1)•DX.

4

Two of the described type of differential equations are called ordinary or abbreviated as ODE. However, the first order equation can be more complex, heterogeneous. They are called LIDE – linear inhomogeneous equation u’ + f(x)•y = g(x).

To LIDE includes, in particular, the Bernoulli equation y’ + f(x)•y = g(x)•y^a. Example: 2•y’ – x2•y = (ln x/X3)•Y2. And the equation in total differentials f(x, y)DX + g(x, y)dy = 0, where ∂FX(x, y)/∂y = ∂do(x, y)/∂X. Example: (X3 – 2•x•y)DX – х2dу = 0, where X3 – 2•x•y – differentiation with respect to x of the function ¼ •x^4 – x2• + C a (x2) – its differentiation with respect to y.

To LIDE includes, in particular, the Bernoulli equation y’ + f(x)•y = g(x)•y^a. Example: 2•y’ – x2•y = (ln x/X3)•Y2. And the equation in total differentials f(x, y)DX + g(x, y)dy = 0, where ∂FX(x, y)/∂y = ∂do(x, y)/∂X. Example: (X3 – 2•x•y)DX – х2dу = 0, where X3 – 2•x•y – differentiation with respect to x of the function ¼ •x^4 – x2• + C a (x2) – its differentiation with respect to y.

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The simplest form of an ODE of second order is y’ + p•a’ + q•u = 0, where p and q are constant coefficients. LIDE of the second order is a complicated version of the ODE, namely u’ + p•a’ + q•y = f(x). Example: y’ – 5•u’ + 13•y = sin x. once p and q are functions of the argument x, the equation might look like this: y’ – 5•x2•y’ + 13•(x - 1)•y = sin X.

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Higher orders differential equations are divided into three subspecies: admitting a reduction of order, equations with constant coefficients and with coefficients as functions of the argument x:

• The expression f(x, y^(m)^(m+1),...,^(n)) = 0 do not contain derivatives of lower order m, then, by replacing z= y^(m) it is possible to reduce the order. Then the equation converts to the form f(x, z, z’,..., z^(n - m)) = 0. Example:’’•x – 4•Y2 = y’ - 2 → z’•x – 4•Y2 = z - 2 where z = y’ = dy/DX;

• LOU u^(k) + p_(k-1)•y^(k-1) + ... + p1•y’ + p0•u = 0 and LIDE have^(k) + p_(k-1)•y^(k-1) + ... + p1•y’ + p0•y = f(x) with constant coefficients pi. Examples: y^(3) + 2•’ – 15•y’ + 3•y = 0 and y^(3) + 2•’ – 15•y’ + 3•y = 2•X3 – ln x;

• LOU u^(k) + p(x)_(k-1)•y^(k-1) + ... + p1(x)•y’ + p0(x)•y = 0 and LIDE have^(k) + p(x)_(k-1)•y^(k-1) + ... + p1(x)•y’ + p0(x)•y = f(x) with coefficients-functions pi(x). Examples: y’’ + 2•x2•’ – 15•agsp x•y’ + 9•x•y = 0 and y’’ + 2•x2•’ – 15•arcsin x•y’ + 9•x•y = 2•X3 – ln Kh.

• The expression f(x, y^(m)^(m+1),...,^(n)) = 0 do not contain derivatives of lower order m, then, by replacing z= y^(m) it is possible to reduce the order. Then the equation converts to the form f(x, z, z’,..., z^(n - m)) = 0. Example:’’•x – 4•Y2 = y’ - 2 → z’•x – 4•Y2 = z - 2 where z = y’ = dy/DX;

• LOU u^(k) + p_(k-1)•y^(k-1) + ... + p1•y’ + p0•u = 0 and LIDE have^(k) + p_(k-1)•y^(k-1) + ... + p1•y’ + p0•y = f(x) with constant coefficients pi. Examples: y^(3) + 2•’ – 15•y’ + 3•y = 0 and y^(3) + 2•’ – 15•y’ + 3•y = 2•X3 – ln x;

• LOU u^(k) + p(x)_(k-1)•y^(k-1) + ... + p1(x)•y’ + p0(x)•y = 0 and LIDE have^(k) + p(x)_(k-1)•y^(k-1) + ... + p1(x)•y’ + p0(x)•y = f(x) with coefficients-functions pi(x). Examples: y’’ + 2•x2•’ – 15•agsp x•y’ + 9•x•y = 0 and y’’ + 2•x2•’ – 15•arcsin x•y’ + 9•x•y = 2•X3 – ln Kh.

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The specific differential equation is not always obvious. Then you should carefully review it to bring to one of the canonical types to apply the appropriate method of solution. This can be done by different methods, the most common ones are replacement and decomposition into components of the derivative y’ = dy/DX.

# Advice 3: How to solve linear equation with two variables

**The equation**, in General written ax+by+C=0, is called a linear equation with two

**variables**. This equation in itself contains an infinite number of solutions, so the task it is always complemented by either another equation or bounding condition. Depending on conditions, will be given a task, solve a linear equation with two

**variables**follows different ways.

You will need

- - linear equation with two variables;
- - the second equation or additional terms.

Instruction

1

If given a system of two linear equations, solve it in the following way. Select one of the equations in which the coefficients in front of

**variables**is smaller and Express one of the variables, e.g., x. Then substitute this value contains, in the second equation. In the resulting equation will have only one variable y, move all the parts from the left side, and the free members to the right. Find and replace any of the original equations to find X.2

To solve a system of two equations in another way. Multiply one of the equations on a number below the coefficient in front of one of the variables, for example, before x was the same in both equations. Then subtract one equation from the other (if the right side is not equal to 0, don't forget to deduct the same and right parts). You will see that the variable x disappeared, leaving only one variable. Solve the resulting equation, and substitute the value found in any of the original equations. Find H.

3

The third method of solving systems of two linear equations – graphical. Draw a coordinate system and draw the graphs of two straight lines whose equations are provided in your system. For this, substitute any two values of x in the equation and find the corresponding y – it will be the coordinates of the points belonging to a straight line. It is most convenient to find the intersection with the coordinate axes is sufficient to substitute the values x=0 and y=0. The coordinates of the point of intersection of these two lines will be the solution of the problem.

4

If the conditions of the problem, only one linear equation, so you are given additional conditions that can help you find a solution. Carefully read the task to find these terms. If

**variables**x and y denoted the distance, speed, age, and weight – feel free to put the constraint x≥0 and y≥0. It is possible, under x or hides the number of children, apples, trees, etc. – then values can only be integers. If x is the age of the son, it is clear that he can't be older than my father, so specify it in terms of the problem.5

Plot the straight line, the corresponding linear equation. Look at the chart, perhaps it will be only a few solutions satisfying all conditions – for example, integers and positive integers. They will be solutions of your equation.

# Advice 4: How to solve equations of higher degrees

The solution to most equations of higher

**degrees**does not have clear formulas such as finding the roots of a quadratic**equation**. However, there are several ways to bring that allow you to convert the equation of higher degree to a more clear mind.Instruction

1

The most common method of solving equations of higher degrees is factorization. This approach is a combination of the selection of the integer roots of the divisors of the free term, and the subsequent division of the total polynomial by binomials of the form (x – x0).

2

For example, solve the equation x^4 + x3 + 2·x2 – x – 3 = 0.Solution.A free member of this polynomial is -3, therefore, its integer divisors can be numbers ±1 and ±3. Substitute them into the equation and see if it will work identity:1: 1 + 1 + 2 – 1 – 3 = 0.

3

*So, the first estimated root gave the correct result. Divide the polynomial equation (x - 1). Polynomial division is performed in a column differs from a conventional division only in the presence of the variable.*

4

Rewrite the equation in a new form (x - 1)·(x3 +2·x2 + 4·x + 3) = 0. The greatest degree of the polynomial was reduced to third. Continue to choose the roots for the cubic polynomial:1: 1 + 2 + 4 + 3 ≠ 0;-1: -1 + 2 – 4 + 3 = 0.

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*The second root x = -1. Divide the cubic polynomial by the expression (x + 1). Record the equation (x - 1)·(x + 1)·(x2 + x + 3) = 0. The degree fell to the second, therefore, the equation can have two roots. To find them, solve the quadratic equation:x2 + x + 3 = 0D = 1 – 12 = -11*

6

The discriminant is negative, then real roots the equation has no more. Find the complex roots of the equation:x = (-2 + i·√11)/2 and x = (-2 – i·√11)/2.

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Write down the answer:x1,2 = ±1; x3,4 = -1/2 ± i·√11/2.

8

Another method for solving equations of higher degree – change of variables to bring it to the square. This approach is used when all of the degree of the equation is even, for example:x^4 – 13·x2 + 36 = 0

9

This equation is called a biquadratic. To bring it to square, make the substitution y = x2. Then:y2 – 13·y + 36 = 0D = 169 – 4·36 = 25y1 = (13 + 5)/2 = 9; y2 = (13 - 5)/2 = 4.

10

Now find the roots of the original equation:x1 = √9 = ±3; x2 = √4 = ±2.