First-Order Linear Equations
Solve the differential equa...

Question

First-Order Linear Equations

Solve the differential equations in Exercises 1–14.

y' + (tanx)y = cos²x, -π/2 < x < π/2

Accepted Answer

Hello. In this video, we are going to be solving for the general solution in the given differential equation. Now the differential equation given to us is Y. Minus tangent of x multiplied by y equal to cosine of x. Now here, if we take a look at our given differential equation. We can see that this matches the form of a linear first order differential equation. Now the general form of a first order linear differential equation is y plus a function of x p x multiplied by y equal to another function of x which will be q x. Here p of x is going to be the function of x in front of our y term, which is going to be negative tangent of x, and q of x is going to be the function of x that the differential equation is equal to, and that is going to be cosine of x. Now, in order to simplify this first order linear differential equation, we need to first solve for our integrating factor. Our integrating factor is going to be defined as m is equal to e raised to the power of the integral of p of x. Now again, we know that P X is going to be negative tangent of X. So our integrating factor is going to equal to E raise the power of negative tangent of X D X. So what we need to do is we need to simplify that integral. Let's just take a look at the integral and the exponential spot for now. So we have the integral of negative tangent of XDX. Now we can simplify this by using the Pythagorean identity for tangent, and we can rewrite the integral to be the integral. The negative integral of sine of x divided by cosine of x. Now here we can simplify by using a u substitution. We will allow u to equal the cosine of x, and that will give us d u is equal to negative sin of x dx, which will further simplify that statement to be negative d u is equal to sin of x d x. So by using this U substitution, we can simplify the integral to be the positive integral of 1 divided by U, DU. This will integrate to be the natural logarithm of U, and once we plug back in our original U in terms of X, that is going to give us the natural logarithm of cosine of X. So that's going to be the simplified form of the exponential integral. So, this is going to simplify our integrating factor to be raised to the power of the natural logarithm of cosine of X, which will further simplify to be cosine of X. So that is going to be the integrating factor to the differential equation. But how do we use this in this integrating factor? Well, now what we will do is we will take our integrating factor, cosine of x, and we are going to multiply to the entirety of our differential equation. This will help us simplify the differential equation further. So by multiplying cosine of x to the entirety of our equation, we will get cosine of x multiplied by Y. minus sin of x. Multiplied by y equal to cosine squared of x. Now if we take a close look at the left hand side of the differential equation, notice that this is equivalent to the derivative of the product cosine of x multiplied by y. So, what we can do is we can rewrite the left hand side of our differential equation as a derivative. So we will have the derivative with respect to x of cosine of x multiplied by y. And this will equal to cosine squared of x. Now, in order to simplify this further, we want to eliminate the differential factor. In order to do this, we will integrate both sides of this equation. Now, the left hand side is going to simplify to be cosine of X multiplied by Y, but now we need to integrate the right-hand side. To simplify the right hand side, we can apply the power reduction identity to rewrite cosine square of x. So we will have the integral of 1 plus cosine of 2 theta divided by 2. This will integrate and simplify to be 1 divided by 2. X Plus 1/4 multiplied by sine of 2 x. And if we were to apply the double angle identity of sin of 2 x, we can further simplify the antiderivative to be 1/2, 1/2 x plus 1/2 multiplied by sin of x multiplied by cosine of x. So this is going to be the integrated form of the right-hand side of the equation. So plugging that back in will give us cosine of x multiplied by y, and that is equal to 1 divided by 2 x. Plus 1 divided by 2 multiplied by sine of x, multiplied by cosine of x, plus the generic constant c. Now, the last thing we want to do is you want to isolate for a Y term. We will divide the entirety of the equation by cosine of x, and that will give us y is equal to 1 divided by 2 cosine of x multiplied by x plus 1 divided by 2 sin of x. Plus the generic constant c divided by cosine of x. And by simplifying the first term by using a Pythagorean identity, the final solution is going to be y is equal to 1 or x divided by 2 multiplied by sequent of x plus 1/2 sin of x plus our generic constant c divided by cosine of x, and this is going to be the final solution to the general solution of the differential equation. And with that being said, the overall solution to this problem is going to be option A. So I hope you, I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

First-Order Linear EquationsSolve the differential equations in Exercises 1–14.y' + (tanx)y = cos²x, -π/2 < x < π/2

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First-Order Linear Equations
Solve the differential equations in Exercises 1–14.

y' + (tanx)y = cos²x, -π/2 < x < π/2