Solve the following differential equation y''+4y=\cos x\sin x | Plainmath

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y^4y y^4y'=x+1. Solve the differential equation.
y y''+4y=sinx solve the differential equation
Differential Equation Variation of Parameters: y Differential Equation Variation of Parameters: y'' + y = ln(sin x)
Differential Equation Undetermined Coefficients: y Differential Equation Undetermined Coefficients: y''' + 4y' = e^x + sin x
FIRST ORDER LINEAR DIFFERENTIAL EQUATION: dy/dx + ycos(x) = sin(x)cos(x) FIRST ORDER LINEAR DIFFERENTIAL EQUATION: dy/dx + ycos(x) = sin(x)cos(x)
Differential Equation Variation of Parameters: y Differential Equation Variation of Parameters: y'' + y = cos x - sin x
Let y = y(x) be the solution of the differential equation 2 cosx dy/dx = sin2x – 4y sinx , x∈(0,π/2) Let y = y(x) be the solution of the differential equation 2 cosx dy/dx = sin2x – 4y sinx , x∈(0,π/2)
1)y 1)y'+ycot x=cos x;(2)y'+y/x=sin x||Step by step solution by using integrating factor for linear ODE.