First Order Differential Equations
2.1 Separable Equations
Separable equations are a type of first-order differential equations that can be rearranged so all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side. This characteristic makes them easier to solve compared to other types of differential equations. Often, these equations represent nonlinear relationships.
Understanding and applying integration techniques is crucial for solving separable equations. Therefore, reviewing and familiarizing yourself with standard integration methods is recommended before attempting to solve these equations.
Solution to Separable Differential Equation
A first-order differential equation is called separable if it can be written in the form of
[asciimath]dy/dx = g(x) p(y)[/asciimath]
where [asciimath]g(x)[/asciimath] is a function of [asciimath]x[/asciimath] only and [asciimath]p(y)[/asciimath] is a function of [asciimath]y[/asciimath] only. The right-hand side is a product of these two functions, allowing the separation of variables.
For example, the equation [asciimath]y'=(x^2+x^2y)/y^2[/asciimath] is separable as it can be factored in and written as [asciimath]dy/dx=x^2((1+y)/y^2)=g(x)p(y)[/asciimath]. However, the equation [asciimath]y'=2-x^2y[/asciimath] is not separable as the right-hand side cannot be factored into a product of the functions of [asciimath]x[/asciimath] and [asciimath]y[/asciimath].
How to Solve Separable Equations
To solve the equation [asciimath]dy/dx = g(x) p(y)[/asciimath],
1. Separate variables: multiply both sides by [asciimath]dx[/asciimath] and by [asciimath]h(y)=1/(p(y))[/asciimath] [asciimath]->[/asciimath] [asciimath]h(y)dy=g(x)dx[/asciimath]
2. Integrate both sides:[asciimath]inth(y)dy=intg(x)dx[/asciimath] [asciimath]->[/asciimath] [asciimath]H(y)=G(x) +C[/asciimath] where [asciimath]C[/asciimath] is the merged constant of integration.
3. Solve for [asciimath]y[/asciimath]: If possible, solve the resulting equation for [asciimath]y[/asciimath] to get the explicit solution. Some solutions cannot be rearranged and solved for [asciimath]y[/asciimath], so the implicit form obtained in Step 2 may be the final solution.
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Solve the nonlinear equation
[asciimath]y'=(x+4)/y^2[/asciimath] .
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1. Multiplying both sides by [asciimath]dx[/asciimath] and [asciimath]y^2[/asciimath] we get
[asciimath]y^2dy=(x+4)dx[/asciimath]
2. Integrating both sides, we get
[asciimath]inty^2dy=int(x+4)dx[/asciimath]
[asciimath]y^3/3=x^2/2+4x+C_1[/asciimath]
3. Multiplying by 3 and taking the cubic root of both sides, we obtain
[asciimath]y=((3x^2)/2+12x+3C_1)^(1//3)[/asciimath]
By substituting constant [asciimath]C_2=3C_1[/asciimath], we’ll have the explicit solution
[asciimath]y=((3x^2)/2+12x+C_2)^(1//3)[/asciimath]
Solve the differential equation
[asciimath]dy/dx =6y tan^2(2x)[/asciimath] .
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This is a separable differential equation as it can be expressed in the form [asciimath]h(y) dy=g(x) dx.[/asciimath]
1. Multiplying both sides by [asciimath]dx[/asciimath] and [asciimath]1/y[/asciimath] we obtain
[asciimath]1/(y)dy = 6tan^2(2x)dx[/asciimath]
2. Integrating both sides, we get
[asciimath]int 1/(y) dy= int 6 tan^2(2 x) dx[/asciimath]
[asciimath]ln|y| = 3tan(2x)-6x+C_1[/asciimath]
3. Exponentiating both sides yields
[asciimath]y= C_2e^(3tan(2x)-6x)[/asciimath] where [asciimath]C_2=e^(C_1)[/asciimath]
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When solving nonlinear differential equations using the separable method, it is crucial to consider the interval of validity, which is the range of the independent variable, typically [asciimath]x[/asciimath], where the solution is defined and behaves appropriately. This interval is essential because solutions to nonlinear equations may not be valid across all [asciimath]x[/asciimath] values due to potential issues like division by zero, undefined logarithms of non-positive numbers, and other undefined operations.
Additionally, due to the nature of nonlinear equations, certain initial conditions might lead to no solution or multiple solutions, emphasizing the need to carefully select and verify the range of [asciimath]x[/asciimath] over which the solution is applicable. The interval of validity is not always immediately apparent from the equation itself and often depends on both the specific form of the solution and the initial conditions.
Solve the initial value problem
[asciimath]y'=14xy-2x[/asciimath] , [asciimath]y(0)=4.[/asciimath]
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Find the general solution:
After factoring out [asciimath]2x[/asciimath] in the right-hand side, the equation can be expressed in the form [asciimath]h(y) dy=g(x) dx.[/asciimath]
[asciimath](dy)/(dx)=2x(7y-1)[/asciimath]
1. Multiplying both sides by [asciimath]dx[/asciimath] and [asciimath]1/(7y-1)[/asciimath] we get
[asciimath]dy/(7y-1)=2xdx[/asciimath]
2. Integrating both sides, we get
[asciimath]intdy/(7y-1) =int2xdx[/asciimath]
[asciimath]1/7ln|7y-1|=x^2+C_1[/asciimath]
3. Multiplying by 7 and exponentiating both sides, we obtain
[asciimath]7y-1=e^(7x^2+7C_1)[/asciimath]
By rearranging the equation and substituting [asciimath]C_2=e^(7C_1)[/asciimath], we’ll have the explicit solution
[asciimath]y=1/7(C_2e^(7x^2)+1)[/asciimath]
Applying the initial condition:
[asciimath]y(0)=4[/asciimath]
[asciimath]1/7(C_2e^0+1)=4[/asciimath]
[asciimath]C_2+1=28[/asciimath]
[asciimath]C_2=27[/asciimath]
The solution to the IVP problem is then
[asciimath]y=1/7(27e^(7x^2)+1)[/asciimath]
There is no restriction on the domain of [asciimath]y[/asciimath], and therefore the solution is valid on [asciimath](-oo,oo)[/asciimath].
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Solve the initial value problem and find the interval of validity of the solution.
[asciimath](dy)/(dx)=(5y^2)/(sqrt(x))[/asciimath] , [asciimath]y(4)=1/35.[/asciimath]
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Find the general solution:
This is a separable differential equation as it can be expressed in the form [asciimath]h(y) dy=g(x) dx.[/asciimath]
1. Multiplying both sides by [asciimath]dx[/asciimath] and [asciimath]1/(y^2)[/asciimath] we get
[asciimath]1/(y^2)\ dy=5/(sqrt(x))\ dx[/asciimath]
2. Integrating both sides, we get
[asciimath]inty^-2\ dy =5intx^(-1/2) \ dx[/asciimath]
[asciimath]-y^-1=10x^(1/2)+C[/asciimath]
3. Multiplying by -1 and taking the reciprocal of both sides, we obtain the explicit solution
[asciimath]y=-1/(10sqrt(x)+C)[/asciimath]
Applying the initial condition:
[asciimath]y(4)=1/35[/asciimath]
[asciimath]-1/(10sqrt(4)+C) =1/35[/asciimath]
[asciimath]-1/(20+C) =1/35[/asciimath]
[asciimath]20+C=-35[/asciimath]
[asciimath]C=-55[/asciimath]
The solution to the IVP problem is then
[asciimath]y=-1/(10sqrt(x)-55)[/asciimath]
Find the interval of validity:
To establish the interval of validity for the solution, we need to consider two constraints:
- The expression within a square root must be positive. Therefore, the term under the square root, should be greater than or equal to 0 ([asciimath]x>=0[/asciimath]).
- The denominator of any rational function should not equal zero to avoid undefined expressions. Given [asciimath]10sqrt(x)-55!=0[/asciimath], it implies that [asciimath]x!=+-30.25[/asciimath].
The interval of validity is the range of [asciimath]x[/asciimath] values that satisfy both conditions:
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Section 2.1 Exercises
- Solve the differential equation: [asciimath]dy/dx =4cos(3x) sqrt(1-y^2)[/asciimath]
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[asciimath]y(x)=sin(4/3sin(3x)+C)[/asciimath]
- Solve the differential equation. Express [asciimath]y[/asciimath] explicitly as a function of [asciimath]x[/asciimath].
[asciimath]dy/dx =4 e^(5x) e^(4y)[/asciimath]
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[asciimath]y(x)=-1/4lnabs(-16/5 e^(5x)+C)[/asciimath]
- Solve the initial value problem: [asciimath]y' = 4x y-2x ,[/asciimath] [asciimath]y(0)=3[/asciimath]
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[asciimath]y(x)=(5 e^(2x^2)+1)/2[/asciimath]
- Solve the initial value problem and find the interval of validity of the solution: [asciimath](dy)/(dx) = (3 y^2)/sqrt(x), \ \ \ y(4) = 1/42[/asciimath]
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[asciimath]y(x)=-1/(6 sqrt(x) -54)[/asciimath]
Interval of validity: [asciimath][0,81)uu(81,oo)[/asciimath]
