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Chapter 4: Problem 25

Solve the differential equation. $$ \frac{d y}{d x}=\frac{x+1}{\left(x^{2}+2 x-3\right)^{2}} $$

Short Answer

Expert verified
The general solution of the given differential equation is \(y = 1/4 \ln|x^2 + 2x -3| - (x^2 + 2x -3) / 2 + C_1\).

Step by step solution

01

Define the Substitution

Let's define a new variable \(u\) such that \(u = x^2 + 2x - 3\). Then, differentiate \(u\) with respect to \(x\) to get \(du/dx = 2x + 2\). This will allow us to substitute \(u\) into the original equation.
02

Substitute into the Differential Equation

Replace \(x^2+2x-3\) with \(u\) and \(x+1\) with \((du/dx - 2) / 2\) in the original differential equation. This transforms the original equation into: \(dy/dx = (du/dx - 2) / (2u^2)\).
03

Simplify the New Differential Equation

Next, multiply both sides of the equation by \(2u^2\) and rearrange terms to isolate \(dy/dx\), which would result in the equivalent form \(dy = (du - 4u) / (4u^2) dx\). This equation is easier to solve with the separation of variables method.
04

Solve with Separation of Variables and Integration

We then separate the variables and integrate both sides. The left side integrates to \(y+C_1\) with \(C_1\) being the constant of integration. The right side is a standard integral which is \(1/4(\ln |u| - 2u)\). After integrating, we then substitute \(u\) back into the equation to get the solution in terms of original variable.
05

Find the General Solution

Substitute the expression for \(u\) back into our integral to find the general solution of the original differential equation. Thus, we have: \(y = 1/4 \ln|x^2 + 2x -3| - (x^2 + 2x -3) / 2+ C_1\). This is the general solution to the given differential equation.

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Most popular questions from this chapter

Find all the continuous positive functions \(f(x),\) for \(0 \leq x \leq\) such that \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} f(x) x d x=\alpha,\) and \(\int_{0}^{1} f(x) x^{2} d x=\alpha^{2}\) where \(\alpha\) is a real number

Show that the function satisfies the differential equation. \(y=a \cosh x\) \(y^{\prime \prime}-y=0\)

In Exercises 83 and \(84,\) use the equation of the tractrix \(y=a \operatorname{sech}^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}, \quad a>0\) Find \(d y / d x\).

Verify the differentiation formula. \(\frac{d}{d x}\left[\operatorname{sech}^{-1} x\right]=\frac{-1}{x \sqrt{1-x^{2}}}\)

Evaluate the integral. \(\int_{0}^{1} \cosh ^{2} x d x\)
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