solve the initial value problems:
1- dq/dz = 2 + sinz, q(0)=5
2-dy/dx = (x^2 -2)x, y(2)=4
3-dy/dx = 4e^5^x + x, y(0)=4
4- dy/dx 3x + 2, y(3)= 2
q = 2 z -cos z + c
when z = 0, q = 5
5 = -1+c
c = 6
so
q = 2z -cos z +6
dy/dx = (x^2 -2)x, y(2)=4
dy/dx = x^3 - 2x
y = (1/4)x^4 - x^2 + c
4 = (1/4)2^4 - 2^2 + c
that should get you started
To solve the initial value problems, we'll use the method of separation of variables and integrate both sides of the differential equation with respect to the respective variables. Let's solve each problem step-by-step:
1) Initial Value Problem: dq/dz = 2 + sin(z), q(0) = 5
Step 1: Separate the variables:
(dq / 2 + sin(z)) = dz
Step 2: Integrate both sides:
∫ dq = ∫ (2 + sin(z)) dz
q = 2z - cos(z) + C
Step 3: Apply the initial condition to find C:
q(0) = 5
5 = 2(0) - cos(0) + C
5 = -1 + C
C = 6
Therefore, the solution to the initial value problem is:
q = 2z - cos(z) + 6
2) Initial Value Problem: dy/dx = (x^2 - 2)x, y(2) = 4
Step 1: Separate the variables:
dy / (x^2 - 2)x = dx
Step 2: Integrate both sides:
∫ dy = ∫ (dx / (x^2 - 2)x)
y = ln(|x^2 - 2|) + C
Step 3: Apply the initial condition to find C:
y(2) = 4
4 = ln(|2^2 - 2|) + C
4 = ln(2) + C
C = 4 - ln(2)
Therefore, the solution to the initial value problem is:
y = ln(|x^2 - 2|) + (4 - ln(2))
3) Initial Value Problem: dy/dx = 4e^(5^x) + x, y(0) = 4
Step 1: Separate the variables:
dy / (4e^(5^x) + x) = dx
Step 2: Integrate both sides:
∫ dy = ∫ (dx / (4e^(5^x) + x))
y = ∫ dx / (4e^(5^x) + x)
Unfortunately, this integral cannot be expressed in terms of elementary functions, so we cannot find a closed-form solution for this problem.
4) Initial Value Problem: dy/dx = 3x + 2, y(3) = 2
Step 1: Separate the variables:
dy / (3x + 2) = dx
Step 2: Integrate both sides:
∫ dy = ∫ dx / (3x + 2)
y = ∫ dx / (3x + 2)
y = (1/3) * ln|3x + 2| + C
Step 3: Apply the initial condition to find C:
y(3) = 2
2 = (1/3) * ln|3(3) + 2| + C
2 = (1/3) * ln|11| + C
C = 2 - (1/3) * ln|11|
Therefore, the solution to the initial value problem is:
y = (1/3) * ln|3x + 2| + 2 - (1/3) * ln|11|
To solve the given initial value problems, we will use the method of separation of variables.
Let's start with the first problem:
1. dq/dz = 2 + sin(z), q(0) = 5
Step 1: Separate the variables.
We have dq = (2 + sin(z)) dz.
Step 2: Integrate both sides with respect to their respective variables.
Integrating dq gives q = 2z - cos(z) + C, where C is the constant of integration.
Step 3: Apply the initial condition to find the value of C.
At z = 0, q = 5. Substituting these values into the equation q = 2z - cos(z) + C gives:
5 = 2(0) - cos(0) + C. Simplifying, we get C = 6.
Step 4: Substitute the value of C back into the equation to find the solution.
The solution to the initial value problem is q = 2z - cos(z) + 6.
Now let's move on to the second problem:
2. dy/dx = (x^2 - 2)x, y(2) = 4
Step 1: Separate the variables.
We have dy = (x^2 - 2)x dx.
Step 2: Integrate both sides with respect to their respective variables.
Integrating dy gives y = (1/4)x^4 - x^2 + C, where C is the constant of integration.
Step 3: Apply the initial condition to find the value of C.
At x = 2, y = 4. Substituting these values into the equation y = (1/4)x^4 - x^2 + C gives:
4 = (1/4)(2^4) - 2^2 + C. Simplifying, we get C = 0.
Step 4: Substitute the value of C back into the equation to find the solution.
The solution to the initial value problem is y = (1/4)x^4 - x^2.
Moving on to the third problem:
3. dy/dx = 4e^(5x) + x, y(0) = 4
Step 1: Separate the variables.
We have dy = (4e^(5x) + x) dx.
Step 2: Integrate both sides with respect to their respective variables.
Integrating dy gives y = (4/5)e^(5x) + (1/2)x^2 + C, where C is the constant of integration.
Step 3: Apply the initial condition to find the value of C.
At x = 0, y = 4. Substituting these values into the equation y = (4/5)e^(5x) + (1/2)x^2 + C gives:
4 = (4/5)e^(5(0)) + (1/2)(0)^2 + C. Simplifying, we get C = 3.
Step 4: Substitute the value of C back into the equation to find the solution.
The solution to the initial value problem is y = (4/5)e^(5x) + (1/2)x^2 + 3.
Lastly, let's solve the fourth problem:
4. dy/dx = 3x + 2, y(3) = 2
Step 1: Separate the variables.
We have dy = (3x + 2) dx.
Step 2: Integrate both sides with respect to their respective variables.
Integrating dy gives y = (3/2)x^2 + 2x + C, where C is the constant of integration.
Step 3: Apply the initial condition to find the value of C.
At x = 3, y = 2. Substituting these values into the equation y = (3/2)x^2 + 2x + C gives:
2 = (3/2)(3)^2 + 2(3) + C. Simplifying, we get C = -20.5.
Step 4: Substitute the value of C back into the equation to find the solution.
The solution to the initial value problem is y = (3/2)x^2 + 2x - 20.5.
That concludes the solutions to the initial value problems.