What is a particular solution to the differential equation #dy/dx=e^(x-y)# with #y(0)=2#?

Question

Charles Anctil · Accepted Answer

# y =ln( e^x + e^2 - 1 )#

this is separable

#dy/dx=e^(x-y) = e^x e^(-y)#

#e^(y)dy/dx= e^x #

#int \ e^(y)dy/dx \ dx=int \ e^x \ dx#

#int \ e^(y) \ dy =int \ e^x \ dx#

# e^(y) = e^x + C#

#y(0) = 2 implies e^2 = 1 + C implies C = e^2 - 1#

# e^(y) = e^x + e^2 - 1#

# y =ln( e^x + e^2 - 1 )#

Christopher Allers · Answer

undefined To find a particular solution to the given differential equation $\frac{dy}{dx} = e^{x-y}$ with the initial condition $y(0) = 2$, we need to solve the differential equation first.

Given:
$$
\frac{dy}{dx} = e^{x-y}
$$

This is a first-order ordinary differential equation. To solve it, we can use the method of separation of variables. Rearranging the equation, we get:
$$
e^y dy = e^x dx
$$

Integrating both sides, we get:
$$
\int e^y dy = \int e^x dx
$$

Solving the integrals, we obtain:
$$
e^y = e^x + C
$$

Where $C$ is the constant of integration. To find $C$, we use the initial condition $y(0) = 2$:
$$
e^2 = e^0 + C \implies C = e^2 - 1
$$

Thus, the solution to the differential equation is:
$$
e^y = e^x + e^2 - 1
$$

To express $y$ explicitly, we take the natural logarithm of both sides:
$$
y = \ln(e^x + e^2 - 1)
$$

Therefore, the particular solution to the differential equation $\frac{dy}{dx} = e^{x-y}$ with the initial condition $y(0) = 2$ is:
$$
y(x) = \ln(e^x + e^2 - 1)
$$