Solve the linear equation cos(x)dxdy(x)+sin(x)y(x)=2cos3(x)sin(x)−1, such that y(4π)=23:Divide both sides by cos(x):dxdy(x)+tan(x)y(x)=−sec(x)+2cos2(x)sin(x)Let μ(x)=e∫tan(x)dx=sec(x).Multiply both sides by μ(x):sec(x)dxdy(x)+(sec(x)tan(x))y(x)=−(sec(x)−2cos2(x)sin(x))sec(x)Substitutetan(x)sec(x)=dxdsec(x):sec(x)dxdy(x)+dxdsec(x)y(x)=−(sec(x)−2cos2(x)sin(x))sec(x)Apply the reverse product rule fdxdg+gdxdf=dxd(fg) to the left-hand side:dxd(sec(x)y(x))=−(sec(x)−2cos2(x)sin(x))sec(x)
Integrate both sides with respect to x :∫dxd(sec(x)y(x))dx=∫−(sec(x)−2cos2(x)sin(x))sec(x)dx Evaluate the integrals:sec(x)y(x)=−21cos(2x)+c1−tan(x), where c1 is an arbitrary constant. Divide both sides by μ(x)=sec(x):y(x)=cos(x)(−21cos(2x)+c1−tan(x)) Solve for c1 using the initial conditions: Substitute y(4π)=23 into y(x)=cos(x)(−21cos(2x)+c1−tan(x)):2c1−1=23 Solve the equation:c1=23+2
Integrate both sides with respect to x :∫dxd(sec(x)y(x))dx=∫−(sec(x)−2cos2(x)sin(x))sec(x)dx Evaluate the integrals: sec(x)y(x)=−21cos(2x)+c1−tan(x), where c1 is an arbitrary constant. Divide both sides by μ(x)=sec(x):y(x)=cos(x)(−21cos(2x)+c1−tan(x)) Solve for c1 using the initial conditions: Substitute y(4π)=23 into y(x)=cos(x)(−21cos(2x)+c1−tan(x)):2c1−1=23Solve the equation:c1=23+2
Substitute c1=23+2 into y(x)=cos(x)(−21cos(2x)+c1−tan(x)) We have the answer to be:y(x)=21cos(x)(−cos(2x)−2tan(x)+2+32)
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