Solved on Feb 22, 2024

Solve the initial value problem: dydx=8xe6y,y(0)=0\frac{d y}{d x}=8 x e^{-6 y}, \quad y(0)=0.

STEP 1

Assumptions
1. The given differential equation is dydx=8xe6y\frac{dy}{dx} = 8xe^{-6y}.
2. The initial condition is y(0)=0y(0) = 0.
3. We are looking for the function y(x)y(x) that satisfies both the differential equation and the initial condition.

STEP 2

To solve the differential equation, we will use the method of separation of variables. This involves rearranging the equation so that all terms involving yy are on one side and all terms involving xx are on the other side.
1e6ydy=8xdx\frac{1}{e^{-6y}}dy = 8x dx

STEP 3

Integrate both sides of the equation with respect to their respective variables.
1e6ydy=8xdx\int \frac{1}{e^{-6y}}dy = \int 8x dx

STEP 4

Simplify the left side of the equation using the property of exponents that e6y=1e6ye^{-6y} = \frac{1}{e^{6y}}.
e6ydy=8xdx\int e^{6y} dy = \int 8x dx

STEP 5

Perform the integration on both sides.
16e6y=4x2+C\frac{1}{6}e^{6y} = 4x^2 + C
Where CC is the constant of integration.

STEP 6

Multiply both sides by 6 to get rid of the fraction.
e6y=24x2+6Ce^{6y} = 24x^2 + 6C

STEP 7

Apply the initial condition y(0)=0y(0) = 0 to find the constant CC.
e60=2402+6Ce^{6 \cdot 0} = 24 \cdot 0^2 + 6C

STEP 8

Simplify the equation using the fact that e0=1e^0 = 1 and 02=00^2 = 0.
1=6C1 = 6C

STEP 9

Solve for CC.
C=16C = \frac{1}{6}

STEP 10

Substitute the value of CC back into the equation.
e6y=24x2+1e^{6y} = 24x^2 + 1

STEP 11

Take the natural logarithm of both sides to solve for yy.
6y=ln(24x2+1)6y = \ln(24x^2 + 1)

STEP 12

Divide both sides by 6 to isolate yy.
y=16ln(24x2+1)y = \frac{1}{6}\ln(24x^2 + 1)
The solution to the initial value problem is y(x)=16ln(24x2+1)y(x) = \frac{1}{6}\ln(24x^2 + 1).

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