Let y=y(x) be the solution of the differential equation
dxdy+2ysec2x=2sec2x+3tanx⋅sec2x, such that y(0)=45. Then 12(y(4π)−e−2) is equal to:
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Step-by-Step Solution
Step 1: Identify the form of the differential equation and find the integrating factor
The given differential equation is of the form of a linear first-order differential equation, which is dxdy+P(x)y=Q(x). Here, P(x)=2sec2x and Q(x)=2sec2x+3tanx⋅sec2x. To solve this, we first need to find the integrating factor (I.F.). The integrating factor is given by the formula e∫P(x)dx.
Step 2: Calculate the integrating factor
We substitute P(x)=2sec2x into the integrating factor formula. The integral of sec2x is tanx. Therefore, the integrating factor is e2tanx.
Step 3: Solve the differential equation
The general solution for a linear first-order differential equation is given by y⋅(I.F.)=∫Q(x)⋅(I.F.)dx. We substitute the expressions for Q(x) and the integrating factor into this formula. To solve the integral on the right-hand side, we can use a substitution.
Step 4: Evaluate the integral using substitution
Let t=2tanx, so dt=2sec2xdx. The integral becomes ∫(1+23t)et2dt. We can split this into two integrals and use integration by parts for the term with tet. After integrating and substituting back t=2tanx, we simplify the expression to find the general solution for y.
Step 5: Apply the initial condition to find C
We are given the initial condition y(0)=45. We substitute x=0 and y=45 into the general solution. Since tan(0)=0 and e0=1, we can solve for the constant C.
Step 6: Find y(4π)
Now that we have the value of C, we can write the particular solution. We need to find y(4π). We substitute x=4π into the particular solution. Since tan(4π)=1, we can simplify the expression.
Step 7: Calculate the final expression
Finally, we need to calculate the value of 12(y(4π)−e−2). We substitute the value of y(4π) we just found and simplify the expression.