Answer to Question #313848 in Differential Equations for Mickey maja

Question

solve the ivp cos(x)y¹ + sin(x)y = 2cos³(x)sin(x) - 1; y(pi/4) =3/2, 0<x<pi/2

Answer 1

$\text{Solve the linear equation } \cos (x) \frac{d y(x)}{d x}+\sin (x) y(x)=2 \cos ^{3}(x) \sin (x)-1 , \text{ such that } y\left(\frac{\pi}{4}\right)=\frac{3}{2} :\\[4mm] \text{Divide both sides by }\cos (x) :\\[2mm] \frac{d y(x)}{d x}+\tan (x) y(x)=-\sec (x)+2 \cos ^{2}(x) \sin (x)\\[2mm] \text{Let }\mu(x)=e^{\int \tan (x) d x}=\sec (x).\\[4mm] \text{Multiply both sides by }\mu(x) :\\[2mm] \sec (x) \frac{d y(x)}{d x}+(\sec (x) \tan (x)) y(x)=-\left(\sec (x)-2 \cos ^{2}(x) \sin (x)\right) \sec (x) Substitute \tan (x) \sec (x)=\frac{d \sec (x)}{d x} :\\ \sec (x) \frac{d y(x)}{d x}+\frac{d \sec (x)}{d x} y(x)=-\left(\sec (x)-2 \cos ^{2}(x) \sin (x)\right) \sec (x)\\[4mm] \text{Apply the reverse product rule }f \frac{d g}{d x}+g \frac{d f}{d x}=\frac{d}{d x}(f g) \text{ to the left-hand side:}\\ \frac{d}{d x}(\sec (x) y(x))=-\left(\sec (x)-2 \cos ^{2}(x) \sin (x)\right) \sec (x)$

$\text{Integrate both sides with respect to x }:\\[2mm] \int \frac{d}{d x}(\sec (x) y(x)) d x=\int-\left(\sec (x)-2 \cos ^{2}(x) \sin (x)\right) \sec (x) d x\\[4mm] \text{ Evaluate the integrals}:\\[2mm] \sec (x) y(x)=-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x), \text{ where } c_{1} \text{ is an arbitrary constant.}\\[4mm] \text{ Divide both sides by }\mu(x)=\sec (x) :\\[2mm] y(x)=\cos (x)\left(-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x)\right)\\[2mm] \text{ Solve for } c_{1} \text{ using the initial conditions:}\\ \text{ Substitute } y\left(\frac{\pi}{4}\right)=\frac{3}{2} \text{ into } y(x)=\cos (x)\left(-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x)\right) :\\[2mm] \frac{c_{1}-1}{\sqrt{2}}=\frac{3}{2}\\[4mm] \text{ Solve the equation:}\\[2mm] c_{1}=\frac{3+\sqrt{2}}{\sqrt{2}}\\$

$\text{ Integrate both sides with respect to x }:\\[2mm] \int \frac{d}{d x}(\sec (x) y(x)) d x=\int-\left(\sec (x)-2 \cos ^{2}(x) \sin (x)\right) \sec (x) d x \\[4mm] \text{ Evaluate the integrals: } \\[2mm] \sec (x) y(x)=-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x), \text{ where } c_{1} \text{ is an arbitrary constant.} \\[4mm] \text{ Divide both sides by } \mu(x)=\sec (x) : \\[2mm] y(x)=\cos (x)\left(-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x)\right)\\[4mm] \text{ Solve for } c_{1} \text{ using the initial conditions: }\\[2mm] \text{ Substitute } y\left(\frac{\pi}{4}\right)=\frac{3}{2} \text{ into } y(x)=\cos (x)\left(-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x)\right) :\\[2mm] \frac{c_{1}-1}{\sqrt{2}}=\frac{3}{2}\\[4mm] \text{Solve the equation:}\\ c_{1}=\frac{3+\sqrt{2}}{\sqrt{2}}\\$

$\text{Substitute } c_{1}=\frac{3+\sqrt{2}}{\sqrt{2}} \text{ into } y(x)=\cos (x)\left(-\frac{1}{2} \cos (2 x)+c_{1}-\tan (x)\right)\\[2mm] \text{ We have the answer to be}:\\ y(x)=\frac{1}{2} \cos (x)(-\cos (2 x)-2 \tan (x)+2+3 \sqrt{2})$