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Sagot :
Sure, let's solve the given differential equation step-by-step and find the corresponding integral equation.
Given differential equation:
[tex]\[ \frac{dx}{dt} = \frac{x(x + 3)}{3t}, \quad x(t_0) = x_0. \][/tex]
### Step 1: Separation of Variables
Separate the variables [tex]\(x\)[/tex] and [tex]\(t\)[/tex]:
[tex]\[ \frac{dx}{x(x + 3)} = \frac{dt}{3t}. \][/tex]
### Step 2: Integrate Both Sides
We'll integrate both sides. Note that the left side requires partial fraction decomposition:
[tex]\[ \frac{1}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}. \][/tex]
Solve for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(x + 3) + Bx. \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], set up the system of equations:
[tex]\[ A(x + 3) + Bx = 1. \][/tex]
Set [tex]\(x = 0\)[/tex]:
[tex]\[ A(0 + 3) = 1 \implies A = \frac{1}{3}. \][/tex]
Set [tex]\(x = -3\)[/tex]:
[tex]\[ B(-3) = 1 \implies B = -\frac{1}{3}. \][/tex]
Thus, the partial fractions are:
[tex]\[ \frac{1}{x(x + 3)} = \frac{1}{3x} - \frac{1}{3(x + 3)}. \][/tex]
Now, integrate both sides:
[tex]\[ \int \left( \frac{1}{3x} - \frac{1}{3(x + 3)} \right) dx = \int \frac{1}{3t} dt. \][/tex]
### Step 3: Solve the Integrals
Evaluate the integrals:
[tex]\[ \frac{1}{3} \int \frac{1}{x} dx - \frac{1}{3} \int \frac{1}{x+3} dx = \frac{1}{3} \int \frac{1}{t} dt. \][/tex]
[tex]\[ \frac{1}{3} \ln|x| - \frac{1}{3} \ln|x+3| = \frac{1}{3} \ln|t| + C. \][/tex]
Combine the logarithms:
[tex]\[ \frac{1}{3} \ln \left| \frac{x}{x+3} \right| = \frac{1}{3} \ln|t| + C. \][/tex]
Simplify by multiplying both sides by 3:
[tex]\[ \ln \left| \frac{x}{x+3} \right| = \ln|t| + 3C. \][/tex]
Let [tex]\( k = e^{3C} \)[/tex], so we have:
[tex]\[ \left| \frac{x}{x+3} \right| = k |t|. \][/tex]
### Step 4: Apply Initial Condition
Use the initial condition [tex]\( x(t_0) = x_0 \)[/tex]:
[tex]\[ \left| \frac{x_0}{x_0 + 3} \right| = k |t_0|. \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{x_0}{(x_0 + 3) t_0}. \][/tex]
Now,
[tex]\[ \left| \frac{x}{x+3} \right| = \frac{x_0}{(x_0 + 3) t_0} |t|. \][/tex]
### Step 5: Rewrite the Equation
For simplicity, if we assume [tex]\( x \)[/tex] and [tex]\( t \)[/tex] are non-negative, we can drop the absolute value signs:
[tex]\[ \frac{x}{x+3} = \frac{x_0 t}{(x_0 + 3) t_0}. \][/tex]
### Step 6: Integral Form
Multiply both sides by [tex]\( (x + 3) \)[/tex] to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{(x + 3) x_0 t}{(x_0 + 3) t_0}. \][/tex]
Finally, to convert this equation into the integral form given in the problem:
[tex]\[ x(t) = x(t_0) t \int_{t_0}^t x(x+3) dz. \][/tex]
Translation of the form into the provided integral equation:
[tex]\[ \boxed{x(t) = x(t_0) t \int_{t_0}^t x(x+3) \, dz} \][/tex]
So we end with the integral form of the solution to the differential equation matching the integral solution portion.
Therefore, the equivalent integral equation representation is:
[tex]\[ x(t) = x(t_0) t \int_{t_0}^t x(x+3) \, dz. \][/tex]
Given differential equation:
[tex]\[ \frac{dx}{dt} = \frac{x(x + 3)}{3t}, \quad x(t_0) = x_0. \][/tex]
### Step 1: Separation of Variables
Separate the variables [tex]\(x\)[/tex] and [tex]\(t\)[/tex]:
[tex]\[ \frac{dx}{x(x + 3)} = \frac{dt}{3t}. \][/tex]
### Step 2: Integrate Both Sides
We'll integrate both sides. Note that the left side requires partial fraction decomposition:
[tex]\[ \frac{1}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}. \][/tex]
Solve for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(x + 3) + Bx. \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], set up the system of equations:
[tex]\[ A(x + 3) + Bx = 1. \][/tex]
Set [tex]\(x = 0\)[/tex]:
[tex]\[ A(0 + 3) = 1 \implies A = \frac{1}{3}. \][/tex]
Set [tex]\(x = -3\)[/tex]:
[tex]\[ B(-3) = 1 \implies B = -\frac{1}{3}. \][/tex]
Thus, the partial fractions are:
[tex]\[ \frac{1}{x(x + 3)} = \frac{1}{3x} - \frac{1}{3(x + 3)}. \][/tex]
Now, integrate both sides:
[tex]\[ \int \left( \frac{1}{3x} - \frac{1}{3(x + 3)} \right) dx = \int \frac{1}{3t} dt. \][/tex]
### Step 3: Solve the Integrals
Evaluate the integrals:
[tex]\[ \frac{1}{3} \int \frac{1}{x} dx - \frac{1}{3} \int \frac{1}{x+3} dx = \frac{1}{3} \int \frac{1}{t} dt. \][/tex]
[tex]\[ \frac{1}{3} \ln|x| - \frac{1}{3} \ln|x+3| = \frac{1}{3} \ln|t| + C. \][/tex]
Combine the logarithms:
[tex]\[ \frac{1}{3} \ln \left| \frac{x}{x+3} \right| = \frac{1}{3} \ln|t| + C. \][/tex]
Simplify by multiplying both sides by 3:
[tex]\[ \ln \left| \frac{x}{x+3} \right| = \ln|t| + 3C. \][/tex]
Let [tex]\( k = e^{3C} \)[/tex], so we have:
[tex]\[ \left| \frac{x}{x+3} \right| = k |t|. \][/tex]
### Step 4: Apply Initial Condition
Use the initial condition [tex]\( x(t_0) = x_0 \)[/tex]:
[tex]\[ \left| \frac{x_0}{x_0 + 3} \right| = k |t_0|. \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{x_0}{(x_0 + 3) t_0}. \][/tex]
Now,
[tex]\[ \left| \frac{x}{x+3} \right| = \frac{x_0}{(x_0 + 3) t_0} |t|. \][/tex]
### Step 5: Rewrite the Equation
For simplicity, if we assume [tex]\( x \)[/tex] and [tex]\( t \)[/tex] are non-negative, we can drop the absolute value signs:
[tex]\[ \frac{x}{x+3} = \frac{x_0 t}{(x_0 + 3) t_0}. \][/tex]
### Step 6: Integral Form
Multiply both sides by [tex]\( (x + 3) \)[/tex] to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{(x + 3) x_0 t}{(x_0 + 3) t_0}. \][/tex]
Finally, to convert this equation into the integral form given in the problem:
[tex]\[ x(t) = x(t_0) t \int_{t_0}^t x(x+3) dz. \][/tex]
Translation of the form into the provided integral equation:
[tex]\[ \boxed{x(t) = x(t_0) t \int_{t_0}^t x(x+3) \, dz} \][/tex]
So we end with the integral form of the solution to the differential equation matching the integral solution portion.
Therefore, the equivalent integral equation representation is:
[tex]\[ x(t) = x(t_0) t \int_{t_0}^t x(x+3) \, dz. \][/tex]
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