Question

In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d u}{d x}=7 x^{6}-3 x^{2}+5\( and \)u=1\( when \)x=1$$

Short answer

The solution to the given initial value problem is \( u(x) = x^7 - x^3 + 5x - 4 \)

Step-by-step solution

Integration

To find \( u(x) \), integrate the given derivative equation with respect to \( x \). The integral of \( \frac{d u}{d x}=7 x^{6}-3 x^{2}+5 \) is \( u(x) = \frac{7x^7}{7} - \frac{3x^3}{3} + 5x + C \) where \( C \) is the unknown arbitrary constant. Simplify to obtain \( u(x) = x^7 - x^3 + 5x + C \)

Apply the Initial Condition

Now apply the given initial condition \( u=1 \) when \( x=1 \) to find the value of the arbitrary constant \( C \). Substituting \( x=1, u=1 \) in the \( u(x) \) equation, we get \( 1 = 1^7 - 1^3 + 5*1 + C \), which simplifies to \( C = 1 \- (1 - 1 + 5) = -4\)

Formulate the Solution

Substitute \( C \) back into the equation of \( u(x) \) to obtain the final solution. Therefore, the solution of the given initial value problem is \( u(x) = x^7 - x^3 + 5x - 4 \)