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Differentiate (dy)/(dx)=e^(x+y) with initial condition
y(0)=-ln 3

Differentiate $\frac{d y}{d x}=e^{x+y}$ with initial condition $y(0)=-\ln 3$

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Find derivatives of radical functions

Full solution

Q. Differentiate

\frac{d y}{d x}=e^{x+y}

with initial condition

y(0)=-\ln 3

Write Differential Equation: Write down the given differential equation and initial condition. $\newline$ $\frac{dy}{dx} = e^{x+y}$ $\newline$ Initial condition: $y(0) = -\ln(3)$
Separate Variables: To solve the differential equation, we need to separate the variables $x$ and $y$ . We can do this by dividing both sides by $e^y$ and multiplying both sides by $dx$ . $\frac{dy}{e^y} = e^x dx$
Integrate Equations: Integrate both sides of the equation with respect to their respective variables. $\newline$ $\int(\frac{1}{e^y}) \, dy = \int e^x \, dx$
Apply Initial Condition: Perform the integration on both sides. $\newline$ The integral of $\frac{1}{e^y}$ with respect to $y$ is $-\frac{1}{e^y}$ , and the integral of $e^x$ with respect to $x$ is $e^x$ . $\newline$ $-\frac{1}{e^y} = e^x + C$ , where $C$ is the constant of integration.
Simplify Equation: Apply the initial condition to find the constant of integration $C$ . $\newline$ $y(0) = -\ln(3)$ implies that when $x = 0$ , $y = -\ln(3)$ . $\newline$ Substitute $x = 0$ and $y = -\ln(3)$ into the integrated equation. $\newline$ $-\frac{1}{e^{-\ln(3)}} = e^0 + C$
Find Constant: Simplify the equation using the properties of exponents and logarithms. $e^{\ln(3)} = 3$ , so $-\frac{1}{3} = 1 + C$ .
General Solution: Solve for $C$ . $C = -\frac{1}{3} - 1$ $C = -\frac{4}{3}$
Solve for y: Write the general solution of the differential equation with the constant $C$ . $-\frac{1}{e^y} = e^x - \frac{4}{3}$
Check for Errors: To express $y$ explicitly in terms of $x$ , we need to solve for $y$ .
$e^y = -\frac{3}{e^x - \frac{4}{3}}$
Taking the natural logarithm of both sides gives us $y$ .
$y = \ln\left(-\frac{3}{e^x - \frac{4}{3}}\right)$
Check for Errors: To express $y$ explicitly in terms of $x$ , we need to solve for $y$ . $e^y = -\frac{3}{e^x - \frac{4}{3}}$ Taking the natural logarithm of both sides gives us $y$ . $y = \ln\left(-\frac{3}{e^x - \frac{4}{3}}\right)$ Check for any mathematical errors in the previous steps, especially in the algebraic manipulations and the application of the initial condition.

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