Let \(y=y(x)\) be the solution of the differential equation \(\frac{dy}{dx}+3\left({\tan }^{2}x\right)y+3y={\sec }^{2}x\…

Equation:

\(\frac{dy}{dx}+3{\tan ⁡}^{2}x\text{ }y+3y={\sec ⁡}^{2}x\)

Note \(3{\tan ⁡}^{2}x+3=3({\tan ⁡}^{2}x+1)=3{\sec ⁡}^{2}x\), so

\(\frac{dy}{dx}+3{\sec ⁡}^{2}x\text{ }y={\sec ⁡}^{2}x\)

Integrating factor:

\(\mu (x)={e}^{\int 3{\sec ⁡}^{2}x\text{ }dx}={e}^{3\tan ⁡x}\)

Thus

\(\frac{d}{dx}(y{e}^{3\tan ⁡x})={\sec ⁡}^{2}x\text{ }{e}^{3\tan ⁡x}\)

Let \(t=\tan ⁡x\), \(dt={\sec ⁡}^{2}x\text{ }dx\):

\(y{e}^{3t}=\int {e}^{3t}dt+C=\frac{1}{3}{e}^{3t}+C\) \(\Rightarrow y=\frac{1}{3}+C{e}^{−3\tan ⁡x}\)

Use \(y(0)=\frac{1}{3}+{e}^{3}\) and \(\tan ⁡0=0\):

\(\frac{1}{3}+C=\frac{1}{3}+{e}^{3}\Rightarrow C={e}^{3}\)

So

\(y(x)=\frac{1}{3}+{e}^{3}{e}^{−3\tan ⁡x}\)

At \(x=\frac{\pi }{4}\), \(\tan ⁡\frac{\pi }{4}=1\):

\(y(\frac{\pi }{4})=\frac{1}{3}+{e}^{3}{e}^{−3}=\frac{1}{3}+1=\frac{4}{3}\)