y′=8e^(4x)(1+y2)solve for y(0)=−2

⇒ y=

I tried to solve the problem and got the equation: tan^-1((2e^(4x)-2+tan(-2)))it says I am wrong and I resolved and got the same thing I must be doing something wrong please help!

dy = 8e^(4x)(1+y^2) dx

dy/(1+y^2) = 8e^(4x) dx
arctan(y) = 2e^(4x)+c

y = tan(2e^(4x)+c)
at x=0 you have
tan(2+c) = -2
c+2 = arctan(-2) = -1.107
c = -3.107

y = tan(2e^(4x)-3.107)

awesome! thank you so much! I see where I went wrong! Would it be ok if I asked you another one I am having trouble with?

To solve the given differential equation y' = 8e^(4x)(1 + y^2) with the initial condition y(0) = -2, we need to find the function y(x).

Let's go through the process step by step:

Step 1: Separate variables
The given differential equation is separable, so we can separate the variables and rewrite it as:

(1 + y^2) dy = 8e^(4x) dx

Step 2: Integrate both sides
Now, we integrate both sides of the equation.

∫(1 + y^2) dy = ∫8e^(4x) dx

The integral on the left-hand side can be simplified to:

y + (1/3) * y^3 = ∫8e^(4x) dx

(Note: We add a constant of integration on both sides, but we'll substitute it later using the initial condition.)

The integral on the right-hand side can be solved using the power rule of integration:

y + (1/3) * y^3 = (8/4) * e^(4x) + C

Simplifying further:

y + (1/3) * y^3 = 2e^(4x) + C

Step 3: Substitute the initial condition
Now, we substitute the initial condition y(0) = -2 into the equation to determine the value of the constant C.

-2 + (1/3) * (-2)^3 = 2e^(4 * 0) + C

-2 - (1/3) * 8 = 2 + C

-2 - (8/3) = 2 + C

-6/3 - 8/3 = 2 + C

-14/3 = 2 + C

C = -14/3 - 2

C = -14/3 - 6/3

C = -20/3

Step 4: Substitute back and simplify
Now, we substitute the value of C back into the equation we obtained earlier:

y + (1/3) * y^3 = 2e^(4x) - 20/3

This is the final solution to the differential equation.

Please note that your solution of tan^-1((2e^(4x)-2+tan(-2))) is incorrect because the given differential equation is not explicitly solvable using trigonometric functions. The correct solution involves both exponential terms and an algebraic expression.