find the general solution of the following equation. express the solution explicitly as a function of the…

Explanation:

Step1: Rewrite the ODE

The given first - order linear ordinary differential equation $u^{\prime}(x)=8e^{9x}-u$ can be rewritten in the standard form $u^{\prime}(x)+u = 8e^{9x}$. The integrating factor is $\mu(x)=e^{\int 1dx}=e^{x}$.

Step2: Multiply through by integrating factor

Multiply the entire equation $u^{\prime}(x)+u = 8e^{9x}$ by $e^{x}$: $e^{x}u^{\prime}(x)+e^{x}u = 8e^{10x}$. The left - hand side is the derivative of the product $e^{x}u$ by the product rule $(uv)^\prime = u^\prime v+uv^\prime$. So, $(e^{x}u)^\prime=8e^{10x}$.

Step3: Integrate both sides

Integrate both sides with respect to $x$: $\int(e^{x}u)^\prime dx=\int 8e^{10x}dx$. The left - hand side is $e^{x}u$, and for the right - hand side, $\int 8e^{10x}dx=8\times\frac{1}{10}e^{10x}+C=\frac{4}{5}e^{10x}+C$. So, $e^{x}u=\frac{4}{5}e^{10x}+C$.

Step4: Solve for $u(x)$

Divide both sides by $e^{x}$ to get $u(x)$: $u(x)=\frac{4}{5}e^{9x}+Ce^{-x}$.

Answer:

$u(x)=\frac{4}{5}e^{9x}+Ce^{-x}$