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Find the solution of the following initial value problems. $$u^{\prime}(x)=\frac{e^{2 x}+4 e^{-x}}{e^{x}} ; u(\ln 2)=2$$

Short Answer

Expert verified
Question: Find the particular solution to the initial value problem (IVP) \(u'(x) = \frac{e^{2x} + 4e^{-x}}{e^x}\) with \(u(\ln 2) = 2\). Answer: The particular solution to the given IVP is \(u(x) = e^x - 2e^{-2x} + \frac{1}{2}\).

Step by step solution

01

Write the equation

Rewrite the given differential equation: $$u'(x) = \frac{e^{2x} + 4e^{-x}}{e^x}$$
02

Integrate both sides

To obtain the general solution, integrate both sides of the equation with respect to x: $$\int u'(x) dx = \int \frac{e^{2x} + 4e^{-x}}{e^x}dx$$
03

Simplify the integrand

Simplify the integrand by dividing the terms inside the integral by \(e^x\): $$\int u'(x) dx = \int \left(e^x + 4e^{-2x}\right)dx$$
04

Integrate the simplified equation

Now, we can easily integrate the simplified equation: $$u(x) = \int \left(e^x + 4e^{-2x}\right)dx = e^x - 2e^{-2x} + C$$ Here, \(C\) is the constant of integration.
05

Apply the initial condition

To find the particular solution, apply the initial condition \(u(\ln 2) = 2\): $$2 = e^{\ln 2} - 2e^{-2\ln 2} + C$$
06

Calculate the constant of integration

Calculate the value of \(C\): $$2 = 2 - \frac{1}{2} + C$$ $$C = \frac{1}{2}$$
07

Write the particular solution

Finally, substitute the value of \(C\) back into the general solution to obtain the particular solution: $$u(x) = e^x - 2e^{-2x} + \frac{1}{2}$$

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Most popular questions from this chapter

The population of a species is given by the function \(P(t)=\frac{K t^{2}}{t^{2}+b},\) where \(t \geq 0\) is measured in years and \(K\) and \(b\) are positive real numbers. a. With \(K=300\) and \(b=30,\) what is \(\lim P(t),\) the carrying capacity of the population? b. With \(K=300\) and \(b=30,\) when does the maximum growth rate occur? c. For arbitrary positive values of \(K\) and \(b,\) when does the maximum growth rate occur (in terms of \(K\) and \(b\) )?

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