A formula that fives prime numbers
Case 3: dydx=fxgySolve the differential equation dydx=(3x-1)(2y+1) given y=4 when x=0Worked solutionExplanationdy2y+1=3x-1.dxSeparate the x and y variables through multiplicative methods only.y=4ydy2y+1=x=0x3x-1.dxIntegrate both sides using the conditions given.124y2.dy2y+1=0x3x-1.dxBalance the integral to form an integral in the form ∫f'yfydy.12ln(2y+1)4y=3x22-x0xEvaluate the integral.12ln|2y+1|-12ln9=3x22-x-(0)12ln|2y+1|=3x22-x+12ln9 ln|2y+1|=3x2-2x+ln9 |2y+1|=e3x2-2x+ln9 2y+1=±e3x2-2x+ln9Simplify and rearrange to make y the subject of the equation.y=12(-1±eln9e3x2-2x)y=129e3x2-2x-1Determine the function by substituting the values x=0 and y=4 into the positive and negative functions and testing which holds true.Solve the differential equation dydx=ex+4y where y=0 when x=0 Worked solutionExplanationdydx=exe4ySplit the exponent using the multiplicative index lawe-4y.dy=ex.dx Separate the x and y variables through multiplicative methods onlyy=0ye-4y.dy=x=0xex.dxIntegrate both sides using the conditions given-e-4y40y=ex0x Evaluate the integral.-e-4y4+14=ex-1Evaluate the integral.-e-4y+1=4ex-4Evaluate the integral.e-4y=5-4exEvaluate the integral.-4y=ln(5-4ex) Simplify. ∴y=-14ln5-4exRearrange to make y the subject of the equation. ................
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