College

Solve [tex]y^{\prime}=-4 \sqrt{x}+24 e^{-6 x}-9 \sin x[/tex], given that the solution passes through the point [tex](0,8)[/tex].

A. [tex]y=-\frac{8}{3} x^{\frac{3}{2}}-4 e^{-6 x}+9 \cos x+3[/tex]

B. [tex]y=-6 x^{\frac{3}{2}}-4 e^{-6 x}-9 \cos x+21[/tex]

C. [tex]y=-6 x^{\frac{3}{2}}-4 e^{-6 x}+9 \cos x+3[/tex]

D. [tex]y=-\frac{8}{3} x^{\frac{3}{2}}-4 e^{-6 x}-9 \cos x+21[/tex]

Answer :

To solve the problem, we need to find the correct function [tex]\( y \)[/tex] that satisfies the given differential equation [tex]\( y' = -4 \sqrt{x} + 24 e^{-6x} - 9 \sin x \)[/tex] and also passes through the point [tex]\((0, 8)\)[/tex].

Here are the steps to identify the correct function:

1. Analyze Candidate Solutions:
We have four candidate solutions provided:

- [tex]\( y_1 = -\frac{8}{3} x^{\frac{3}{2}} - 4 e^{-6x} + 9 \cos x + 3 \)[/tex]
- [tex]\( y_2 = -6 x^{\frac{3}{2}} - 4 e^{-6x} - 9 \cos x + 21 \)[/tex]
- [tex]\( y_3 = -6 x^{\frac{3}{2}} - 4 e^{-6x} + 9 \cos x + 3 \)[/tex]
- [tex]\( y_4 = -\frac{8}{3} x^{\frac{3}{2}} - 4 e^{-6x} - 9 \cos x + 21 \)[/tex]

2. Check Passing Through the Point (0, 8):
For each candidate solution, substitute [tex]\( x = 0 \)[/tex] and see if [tex]\( y = 8 \)[/tex].

- For [tex]\( y_1 \)[/tex], substitute [tex]\( x = 0 \)[/tex]:

[tex]\( y_1 = -\frac{8}{3} \cdot 0^{\frac{3}{2}} - 4 e^{-6 \cdot 0} + 9 \cos(0) + 3 \)[/tex]

Simplifying gives:

[tex]\( y_1 = 0 - 4 \cdot 1 + 9 \cdot 1 + 3 = 0 - 4 + 9 + 3 = 8 \)[/tex]

Since this satisfies the point [tex]\((0, 8)\)[/tex], [tex]\( y_1 \)[/tex] could be the solution.

3. Verify Differential Equation:
Finally, we need to check if the derivative of this [tex]\( y_1 \)[/tex], denoted as [tex]\( y_1' \)[/tex], matches the given differential equation:

- Calculate [tex]\( y_1' \)[/tex]:

- The derivative of [tex]\(-\frac{8}{3} x^{\frac{3}{2}}\)[/tex] is [tex]\(-4\sqrt{x}\)[/tex].
- The derivative of [tex]\(-4 e^{-6x}\)[/tex] is [tex]\(24 e^{-6x}\)[/tex].
- The derivative of [tex]\(9 \cos x\)[/tex] is [tex]\(-9 \sin x\)[/tex].
- The derivative of the constant 3 is 0.

- Thus, [tex]\( y_1' = -4 \sqrt{x} + 24 e^{-6x} - 9 \sin x \)[/tex], which matches the given differential equation.

Since [tex]\( y_1 \)[/tex] passes through the point [tex]\((0, 8)\)[/tex] and its derivative matches the given expression for [tex]\( y'\)[/tex], the correct solution is:

[tex]\( y = -\frac{8}{3} x^{\frac{3}{2}} - 4 e^{-6x} + 9 \cos x + 3 \)[/tex].