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.

5

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 an equation in mathematics

The word "

**equation**" suggests that recorded kind of equality. There are known and unknowns. There are different types of equations - logarithmic, exponential, trigonometric and others. Consider how to learn how to solve equations, example of linear equations.Instruction

1

Learn how to solve a simple linear

**equation**of the form ax+b=0. x is an unknown that must be found. Are called linear equations in which x can only be in the first degree, no squares and cubes. a and b are any numbers and a cannot equal 0. If a or b are represented as fractions, the denominator is never x. Otherwise you can get a linear**equation**.Solved linear**equation**simple. Move b to the other side of the equal sign. The sign that stood in front of b, is reversed. Was a plus - will be negative. Get ax=-b.Now find x, which divide both parts of equality on a. We get x=-b/a.2

To solve more complex equations, remember the 1st identity transformation. Its meaning is the following. To both sides of this equation we can add the same number or expression. And by analogy, from both sides of the equation, you can subtract the same number or expression.Suppose you have

**the equation**5x+4=8. Subtract from the left and right side is the same expression (5x+4). We get 5x+4-(5x+4)=8-(5x+4). After opening parentheses is 5x+4-5x-4=8-5x-4. The result is 0=4-5x. Looks at this**equation**differently, but its essence remains the same. The initial and final equations are called identically equal.3

Remember 2-d identity transformation. Both parts of the equation can be multiplied by same number or expression. By analogy - both parts of the equation can be divided into the same number or expression. Naturally, you should not multiply or divide by 0.Let there be

**the equation**1=8/(5x+4). Multiply both sides by the same expression (5x+4). We get 1*(5x+4)=(8*(5x+4))/(5x+4). After the reduction we get 5x+4=8.4

Learn with the help of simplifications and transformations allow the linear equations to a familiar sight. Let there be

**the equation**(2x+4)/3-(5x-2)/2=11+(x-4)/6. This**equation**is exactly linear, because x is in the first degree and in the denominators of fractions x is missing. But**the equation is**not like a simple, parsed into 1-m step.Apply 2-d identity transformation. Multiply both sides by the number 6 is the common denominator of all the fractions. We get 6*(2x+4)/3-6*(5x-2)/2=6*11+6*(x-4)/6. After the reduction of the numerator and denominator have 2*(2x+4)-3*(5x-2)=66+1*(x-4). The choir 4x+8-15x+6=66+x-4. 14-11x=62+x.Applicable 1-e the identity transformation. Subtract from the left and right side of the expression (62+x). Get 14-11x-(62+x)=62+x-(62+x). 14-11x-62-x=0. We get-12x-48=0. And it is the simplest linear**equation**, the solution of which dismantled at the 1st step. Complex initial expression with fractions we have presented in the usual form, using the identity transformation.Note

Often mistakes are made when disclosure of brackets. Remember that if before the parenthesis is a minus sign, while getting rid of the brackets, the signs change to the opposite. For example, at the 4th step of opening bracket -(62+x)=-62-x.

Useful advice

Solve more equations textbook at the end of which there are answers. Check the correctness of assignments.

# Advice 4: 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 5: How to solve differential equation

Tasks on the differential and integral calculus are important elements of consolidation theory, mathematical analysis, section of mathematics that is studied in universities. The differential

**equation**is solved by integration.Instruction

1

Differential calculus explores the properties of functions. Conversely, the integration of functions allows for these properties, i.e. derivatives or differentials of functions to find herself. This is the solution of the differential equation.

2

Every equation is a relation between the unknown variable and known data. In the case of differential equations plays the role of the unknown function, and the role is known to value its derivatives. In addition, the ratio can contain the independent variable:F(x, y(x), y’(x), y’(x),..., y^n(x)) = 0, where x is an unknown variable, y(x) is the function you want to define the order of an equation is the maximum order of derivative (n).

3

This equation is called an ordinary differential equation. If the correlation of multiple independent variables and partial derivatives (differentials) of functions in these variables, the equation is called a differential equation with partial derivatives and has the form:x∂z/∂y - ∂z/∂x = 0, where z(x, y) is the desired function.

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So, to learn how to solve differential equations, you should be able to find the integral, i.e., to solve the problem opposite of differentiation. Example:Solve the equation of first order y’ = -y/x.

5

Will resentement y’ for dy/dx: dy/dx = -y/x.

6

Give the equation to the form convenient for integration. To do this, multiply both sides by dx and divide by y:dy/y = -dx/x.

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Integrate:∫dy/y = - ∫dx/x + Сln |y| = - ln |x| + C.

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Imagine a constant of the natural logarithm, C = ln |C|, then:ln|xy| = ln|C|, where xy = C.

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This solution is called the General solution of the differential equation. C is a constant, a set of values which defines the set of solutions of the equation. At any particular value of the decision would be the one. This solution is a particular solution of the differential equation.

# Advice 6: 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.

5

*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*

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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.

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