91Ó°ÊÓ

Integrate both sides of the ODE with respect to \(\theta\): $$\int y^{\prime}(\theta)d\theta=\int\frac{\sqrt{2} \cos ^{3} \theta+1}{\cos ^{2} \theta}d\theta$$ Note that \(\cos^{2} \theta = 1 - \sin^{2} \theta\). Let \(u = \sin\theta\), then: $$du = \cos\theta d\theta$$ Now, substitute \(u\) in the integral: $$\int y^{\prime}(\theta)d\theta=\int\frac{\sqrt{2} (1-u^{2})^{\frac{3}{2}}+1}{1-u^{2}} du$$ Now integrate the right-hand side: $$y(\theta) = \int\frac{\sqrt{2} (1-u^{2})^{\frac{3}{2}}+1}{1-u^{2}} du+C$$ Replace \(u\) back with \(\sin \theta\): $$y(\theta) = \int\frac{\sqrt{2} (1-\sin^{2} \theta)^{\frac{3}{2}}+1}{1-\sin^{2} \theta} d\sin\theta+C$$

Now, we will use the initial condition, \(y\left(\frac{\pi}{4}\right)=3\), to find the constant of integration \(C\). Plug \(\theta = \frac{\pi}{4}\) and \(y(\theta) =3\) into the antiderivative formula and solve for \(C\): $$3 = \int\frac{\sqrt{2} (1-\sin^{2} \left(\frac{\pi}{4}\right))^{\frac{3}{2}}+1}{1-\sin^{2} \left(\frac{\pi}{4}\right)} d\sin\left(\frac{\pi}{4}\right)+C$$ We know that \(\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\). Substitute it and simplify the equation: $$3 = \int\frac{\sqrt{2} (1-\frac{1}{2})^{\frac{3}{2}}+1}{1-\frac{1}{2}} d\left(\frac{1}{\sqrt{2}}\right)+C$$ $$3 = \int\frac{\sqrt{2} \left(\frac{1}{2}\right)^{\frac{3}{2}}+1}{\frac{1}{2}} d\left(\frac{1}{\sqrt{2}}\right)+C$$ $$3 = 2 (\sqrt{2} (1)+1) \frac{1}{\sqrt{2}}+C$$ Therefore, $$C = 3-2(\sqrt{2}+1)$$

Substitute the value of \(C\) in the antiderivative formula to get the general solution of the initial value problem: $$y(\theta) = \int\frac{\sqrt{2} (1-\sin^{2} \theta)^{\frac{3}{2}}+1}{1-\sin^{2} \theta} d\sin\theta+3-2(\sqrt{2}+1)$$ This is the solution of the initial value problem: $$y(\theta) = \int\frac{\sqrt{2} (1-\sin^{2} \theta)^{\frac{3}{2}}+1}{1-\sin^{2} \theta} d\sin\theta+3-2\sqrt{2}-2$$