9. $[\mathbf{7}]$ Let $x(t)$ be a solution to the differential equation
$$
\left(x+x^{\prime}\right)^{2}+x \cdot x^{\prime \prime}=\cos t
$$
with $x(0)=x^{\prime}(0)=\sqrt{\frac{2}{5}}$. Compute $x\left(\frac{\pi}{4}\right)$.
the equation becomes $y+y^{\prime}+\frac{1}{2} y^{\prime \prime}=\cos t$. The term $\cos t$ suggests that the particular solution should be in the form $A \sin t+B \cos t$. By substitution and coefficient comparison, we get $A=\frac{4}{5}$ and $B=\frac{2}{5}$. Since the function $y(t)=\frac{4}{5} \sin t+\frac{2}{5} \cos t$ already satisfies the initial conditions $y(0)=x(0)^{2}=\frac{2}{5}$ and $y^{\prime}(0)=2 x(0) x^{\prime}(0)=\frac{4}{5}$, the function $y$ also solves the initial value problem. Note that since $x$ is positive at $t=0$ and $y=x^{2}$ never reaches zero before $t$ reaches $\frac{\pi}{4}$, the value of $x\left(\frac{\pi}{4}\right)$ must be positive. Therefore, $x\left(\frac{\pi}{4}\right)=+\sqrt{y\left(\frac{\pi}{4}\right)}=\sqrt{\frac{6}{5} \cdot \frac{\sqrt{2}}{2}}=\frac{\sqrt[4]{450}}{5}$.