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Chapter 20: Problem 7

Integrate the equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^{2}\) subject to the condition \(y(1)=4\) in order to find the particular solution.

Short Answer

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Question: Find the particular solution of the given differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^{2}\) with the initial condition \(y(1)=4\). Answer: The particular solution of the given differential equation is \(y(x) = x^{3} + 3\).

Step by step solution

01

Write the given differential equation

We are given the differential equation: $$\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^{2}$$
02

Integrate the equation with respect to \(x\)

To find the solution \(y(x)\), we need to integrate both sides with respect to \(x\): $$\int \frac{\mathrm{d} y}{\mathrm{~d} x} \mathrm{~d} x = \int 3 x^{2} \mathrm{~d} x$$ Applying the integration, we get: $$y(x) = x^{3} + C$$ where C is the integration constant.
03

Use the initial condition to find the value of the constant \(C\)

We have the initial condition: \(y(1)=4\). Plug this into the equation to solve for \(C\): $$4 = 1^{3} + C$$ Solving for \(C\), we get \(C = 3\).
04

Write down the particular solution

Now that we have the value of the constant \(C\), we can write the particular solution of the given differential equation: $$y(x) = x^{3} + 3$$

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Most popular questions from this chapter

Find a first-order equation satisfied by \(x=A \mathrm{e}^{-2 x}\),

Find a second-order differential equation that is satisfied by $$ y=A \cosh 2 x+B \sinh 2 x $$

Using software, obtain a symbolic solution of \(i R+L \frac{\mathrm{d} i}{\mathrm{~d} t}=E\) where \(R, L\) and \(E\) are constants subject to the initial condition \(i(0)=0\).

Use Euler's method to find a numerical solution of \(\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{x y}{x^{2}+2}\) subject to \(y(1)=3\). Take \(h=0.1\) and hence approximate \(y(1.5)\). Obtain the true solution using the method of separation of variables. Work throughout to six decimal places.

Use a package to find the general solution of \(\frac{\mathrm{d} T}{\mathrm{~d} \theta}=\mu(T-K)\) where \(\mu\) and \(K\) are constants.
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