Let y=y(x)y = y(x) be the solution of the differential equation dydx+2ysec⁡2x=2sec⁡2x+3tan⁡x⋅sec⁡2x\frac{dy}{dx} + 2y \sec^2 x = 2\sec^2 x + 3 \tan x \cdot \sec^2 x, such that y(0)=54y(0) = \frac{5}{4}. Then 12(y(π4)−e−2)12 \left( y\left(\frac{\pi}{4}\right) - e^{-2} \right) is equal to:

Answer: 15+6e−215 + 6 e^{-2}

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 dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x). Here, P(x)=2sec⁡2xP(x) = 2\sec^2 x and Q(x)=2sec⁡2x+3tan⁡x⋅sec⁡2xQ(x) = 2\sec^2 x + 3\tan x \cdot \sec^2 x. To solve this, we first need to find the integrating factor (I.F.). The integrating factor is given by the formula e∫P(x)dxe^{\int P(x) dx}.

Step 2: Calculate the integrating factor

We substitute P(x)=2sec⁡2xP(x) = 2\sec^2 x into the integrating factor formula. The integral of sec⁡2x\sec^2 x is tan⁡x\tan x. Therefore, the integrating factor is e2tan⁡xe^{2\tan x}.

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.)dxy \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx. We substitute the expressions for Q(x)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=2tan⁡xt = 2\tan x, so dt=2sec⁡2xdxdt = 2\sec^2 x dx. The integral becomes ∫(1+32t)etdt2\int (1 + \frac{3}{2}t) e^t \frac{dt}{2}. We can split this into two integrals and use integration by parts for the term with tett e^t. After integrating and substituting back t=2tan⁡xt = 2\tan x, we simplify the expression to find the general solution for yy.

Step 5: Apply the initial condition to find C

We are given the initial condition y(0)=54y(0) = \frac{5}{4}. We substitute x=0x=0 and y=54y=\frac{5}{4} into the general solution. Since tan⁡(0)=0\tan(0) = 0 and e0=1e^0 = 1, we can solve for the constant CC.

Step 6: Find y(π4)y\left(\frac{\pi}{4}\right)

Now that we have the value of CC, we can write the particular solution. We need to find y(π4)y\left(\frac{\pi}{4}\right). We substitute x=π4x = \frac{\pi}{4} into the particular solution. Since tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1, we can simplify the expression.

Step 7: Calculate the final expression

Finally, we need to calculate the value of 12(y(π4)−e−2)12 \left( y\left(\frac{\pi}{4}\right) - e^{-2} \right). We substitute the value of y(π4)y\left(\frac{\pi}{4}\right) we just found and simplify the expression.

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