y′=x−2ycot⁡2x

Name: y′=x−2ycot⁡2x
Uploaded: 2022-02-23T17:22:57+00:00
Description: y′=x−2ycot⁡2x

Question

y′=x−2ycot⁡2x

Piosellisf · Accepted Answer

Step 1Given that the equation y′=x−2ycot⁡2xy′+(2cot⁡2x)y=xThis is a linear differential equation in y.dydx+Py=QHere, P=2cot⁡2x,Q=x∫Pdx=∫2cot⁡2xdx=2[log⁡sin⁡2x2]=log⁡sin⁡2xThus, integral factor, I.F.=e∫P dx=elog⁡sin⁡2x=sin⁡2xStep 2Hence the solution isye∫ p dx=∫Qe∫ p dxdx+Cy(sin⁡2x)=∫x(sin⁡2x)dx+Cy(sin⁡2x)=[x(−cos⁡2x2)+(1)(−sin⁡2x4)]+Cy(sin⁡2x)=[−xcos⁡2x2−sin⁡2x4]+Cysin⁡2x=14(−2xcos⁡2x−sin⁡2x)+C is the required solution.

ol3i4c5s4hr · Accepted Answer

y′=x−2ycot⁡(2x)y′+2ycot⁡(2x)=xBy inspection, the equation is a linear DE in the form of:y′+yP(x)=Q(x)where: P(x)=2cot⁡(2x) and Q(x)=xThe integrating factor, i.f. is:i.f.=e∫P(x)dxi.f=e∫2cot⁡(2x)dxNote: ∫cot⁡u du=ln⁡(sin⁡u)+Ci.f.=eln⁡(sin⁡(2x))Note: eln⁡u=ui.f=sin⁡(2x)Substituting the i.f., we get:ysin⁡(2x)=∫×sin⁡(2x)dx+CFor ∫×sin⁡(2x)dx, use integration by parts.In integration by parts, choose u in this order: LIATELogarithmicInverseAlgebraicTrigonometricExponential∫×sin⁡(2x)dxUsing integration by parts:Let: u=xdu=dxdv=sin⁡(2x)dxv=−12cos⁡(2x)∫udv=uv−∫vdu∫×sin⁡(2x)dx=−12×cos⁡(2x)−∫−12cos⁡(2x)dx∫×sin⁡(2x)dx=−12×cos⁡(2x)+14sin⁡(2x)Substituting the value of ∫×sin⁡(2x)dx, we get:

Vasquez · Accepted Answer

Simplifying y=x+-2y * cot * 2x Reorder the terms for easier multiplication: y=x+-2 * 2y * cot * x Multiply -2 * 2 Multiply y * cot Multiply cot y * x Reorder the terms: y=-4 cot xy+x Solving y=-4 cot xy+x Solving for variable &quot;y&quot;. Move all terms containing y to the left, all other terms to the right. Add &quot;4 cot xy&quot; to each side of the equation. 4 cot xy+y=-4 cot xy+4 cot xy+x Combine like terms: -4 cot xy+4 cot xy=0 4 cot xy+y=0+x 4 cot xy+y=x Reorder the terms: 4 cot xy+-1x+y=x+-1x Combine like terms: x+-1x= 4 cot xy+-1x+y=0 The solution to this equation could not be determined.