Math Problem Statement

Find y as a function of x if its rate of change and boundary conditions are given below:

dydx= (x3+30x−102)4(x2+10) , y=50,663 when x=3 .

Answer y as a function of x (keep the antiderivative in factored form):

y=

Solution

We are tasked with finding yy as a function of xx, given the following rate of change and boundary conditions:

dydx=(x3+30x102)4(x2+10)\frac{dy}{dx} = \frac{(x^3 + 30x - 102)}{4(x^2 + 10)} with the boundary condition y(3)=50,663y(3) = 50,663.

Step 1: Integrate the rate of change

To find y(x)y(x), we need to integrate the derivative dydx\frac{dy}{dx}:

y(x)=(x3+30x102)4(x2+10)dxy(x) = \int \frac{(x^3 + 30x - 102)}{4(x^2 + 10)} dx

We can split the integrand into two parts for easier integration. Let’s divide:

dydx=x34(x2+10)+30x4(x2+10)1024(x2+10)\frac{dy}{dx} = \frac{x^3}{4(x^2 + 10)} + \frac{30x}{4(x^2 + 10)} - \frac{102}{4(x^2 + 10)}

1. First term: x34(x2+10)\frac{x^3}{4(x^2 + 10)}

We can rewrite it as:

x3x2+10=x10xx2+10\frac{x^3}{x^2 + 10} = x - \frac{10x}{x^2 + 10}

So, the integral of the first term becomes:

x34(x2+10)dx=14(x10xx2+10)dx=14(x225ln(x2+10))\int \frac{x^3}{4(x^2 + 10)} dx = \frac{1}{4} \int \left( x - \frac{10x}{x^2 + 10} \right) dx = \frac{1}{4} \left( \frac{x^2}{2} - 5 \ln(x^2 + 10) \right)

2. Second term: 30x4(x2+10)\frac{30x}{4(x^2 + 10)}

This is a standard logarithmic integral:

30x4(x2+10)dx=152ln(x2+10)\int \frac{30x}{4(x^2 + 10)} dx = \frac{15}{2} \ln(x^2 + 10)

3. Third term: 1024(x2+10)- \frac{102}{4(x^2 + 10)}

This can be integrated as:

1024(x2+10)dx=512arctan(x10)\int \frac{-102}{4(x^2 + 10)} dx = - \frac{51}{2} \arctan\left(\frac{x}{\sqrt{10}}\right)

Step 2: General solution

Combining these, we get the general form of y(x)y(x) (in factored form) as:

y(x)=18x254ln(x2+10)+152ln(x2+10)512arctan(x10)+Cy(x) = \frac{1}{8}x^2 - \frac{5}{4} \ln(x^2 + 10) + \frac{15}{2} \ln(x^2 + 10) - \frac{51}{2} \arctan\left(\frac{x}{\sqrt{10}}\right) + C

Simplifying the logarithmic terms:

y(x)=18x2+254ln(x2+10)512arctan(x10)+Cy(x) = \frac{1}{8}x^2 + \frac{25}{4} \ln(x^2 + 10) - \frac{51}{2} \arctan\left(\frac{x}{\sqrt{10}}\right) + C

Step 3: Apply boundary condition

We are given that y(3)=50,663y(3) = 50,663. Plugging x=3x = 3 into the equation:

50,663=18(32)+254ln(32+10)512arctan(310)+C50,663 = \frac{1}{8}(3^2) + \frac{25}{4} \ln(3^2 + 10) - \frac{51}{2} \arctan\left(\frac{3}{\sqrt{10}}\right) + C

Solving this will give us the value of CC.

Let's compute that constant.The constant CC is approximately 50,662.8350,662.83.

Final solution:

Thus, the function y(x)y(x) is:

y(x)=18x2+254ln(x2+10)512arctan(x10)+50,662.83y(x) = \frac{1}{8}x^2 + \frac{25}{4} \ln(x^2 + 10) - \frac{51}{2} \arctan\left(\frac{x}{\sqrt{10}}\right) + 50,662.83

Would you like further details on any part of the solution or have additional questions?


Here are 5 related questions for deeper exploration:

  1. How does the logarithmic term affect the behavior of y(x)y(x) as xx \to \infty?
  2. What is the geometric interpretation of the arctangent term in this solution?
  3. How can this technique be applied to similar integrals involving rational functions?
  4. How would the solution change if a different boundary condition was provided?
  5. What are the practical applications of solving such differential equations?

Tip: When integrating, breaking complex terms into simpler parts often simplifies the calculation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Integration
Logarithmic and Trigonometric Functions

Formulas

dy/dx = (x^3 + 30x - 102) / 4(x^2 + 10)
y(x) = ∫ (x^3 + 30x - 102) / 4(x^2 + 10) dx
ln(x^2 + 10)
arctan(x/√10)

Theorems

Fundamental Theorem of Calculus
Logarithmic and Trigonometric Integration

Suitable Grade Level

College Level