Let y=y(x) be the solution of the differential equation cosx(loge(cosx))2dy+(sinx−3ysinxloge(cosx))dx=0,x∈(0,2π). If y(4π)=loge2−1, then y(6π) is equal to:
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Step-by-Step Solution
Step 1: Rearrange the differential equation
First, we need to rearrange the given differential equation into the standard form of a linear first-order differential equation, which is dxdy+P(x)y=Q(x). This involves dividing by dx and isolating the terms involving y and constants.
Step 2: Identify P(x) and Q(x)
From the rearranged equation, we can identify the functions P(x) and Q(x). These functions are crucial for finding the integrating factor.
Step 3: Calculate the integrating factor
The integrating factor (I.F.) is calculated using the formula e∫P(x)dx. We perform a substitution u=loge(cosx), so du=cosx1(−sinx)dx=−tanxdx. This simplifies the integral, leading to the integrating factor (loge(cosx))−3.
Step 4: Solve the differential equation
The general solution is given by y⋅I.F.=∫Q(x)⋅I.F.dx. We substitute the expressions for Q(x) and the I.F. and integrate. Again, using the substitution u=loge(cosx), the integral becomes straightforward, yielding the general solution for y(x).
Step 5: Apply the initial condition to find C
We use the given initial condition y(4π)=loge2−1 to find the value of the constant C. Substituting x=4π into the general solution and using cos(4π)=21, we find that C=0.
Step 6: Calculate y(6π)
With C=0, the particular solution is y(x)=21(loge(cosx))−2. Now, we substitute x=6π into this solution. Since cos(6π)=23, we can calculate the final value of y(6π).