Problem 43
Question
Solve the given differential equation. $$2 x\left(y^{\prime}+y^{3} x^{2}\right)+y=0$$
Step-by-Step Solution
Verified Answer
The given differential equation is \(2x(y'(x) + y^3(x^2))+y(x)=0\). After attempting to rewrite and simplify the equation, we performed a substitution with \(v(x) = y^2(x)\) and found a new differential equation \(\frac{1}{2}v'(x) + v(x^2) = -\frac{\sqrt{v(x)}}{2x}\). This new equation is still nonlinear and might require more advanced techniques to solve.
1Step 1: Rewrite the given differential equation in a more familiar form.
First, let's rewrite the given differential equation in terms of y(x) and its derivative y'(x), as follows:
\(2x(y'(x) + y^3(x^2))+y(x)=0\)
Now, let's isolate the derivative term, y'(x), on the left side of the equation:
\(y'(x)+y^3(x^2) = -\frac{y(x)}{2x}\)
2Step 2: Check if the equation is an exact or separable differential equation.
Unfortunately, this differential equation doesn't seem to be exact or separable. Therefore, we will try a different approach to solving the equation.
3Step 3: Attempt a substitution.
Let's make a substitution to simplify the equation. We can substitute:
\(v(x) = y^2(x)\)
Then, the derivative of v(x) with respect to x would be:
\(v'(x) = 2y(x)y'(x)\)
Now, let's substitute v(x) into the original differential equation:
\(y'(x) + y^3(x^2) = -\frac{y(x)}{2x}\)
Substituting y^2(x) with v(x) and 2y(x)y'(x) with v'(x):
\(\frac{1}{2}v'(x) + v(x^2) = -\frac{\sqrt{v(x)}}{2x}\)
4Step 4: Analyze the new equation.
Now we have a new differential equation:
\(\frac{1}{2}v'(x) + v(x^2) = -\frac{\sqrt{v(x)}}{2x}\)
Although this differential equation is still nonlinear, it is simpler than the original one. We can now attempt to solve this equation, but it most likely requires more advanced techniques beyond the scope of a high school course.
Key Concepts
Nonlinear Differential EquationsSubstitution Method in Differential EquationsSolving Differential Equations Step by Step
Nonlinear Differential Equations
When students begin to solve differential equations, they often start with linear differential equations, which have solutions that can be added together to form new solutions—a property known as superposition. But in the world of mathematics, not all equations play by such simple rules; this is where nonlinear differential equations come into play.
Unlike their linear counterparts, nonlinear differential equations involve terms that are not simply proportional to the function and its derivatives; they may include powers or products of the function and its derivatives. For instance, an equation like
\[y'(x) + y^3(x^2) = -\frac{y(x)}{2x}\]
includes a cubic term of the dependent variable, making it nonlinear. These equations are known for modeling more complex phenomena in physics and biology but are also notoriously harder to solve.
Nonlinear equations do not generally have straightforward solutions and can exhibit behaviors such as sensitivity to initial conditions, making them challenging yet fascinating to study. They often require specific methods for finding solutions, such as perturbation techniques, numerical solutions, or, as in our exercise, the substitution method.
Unlike their linear counterparts, nonlinear differential equations involve terms that are not simply proportional to the function and its derivatives; they may include powers or products of the function and its derivatives. For instance, an equation like
\[y'(x) + y^3(x^2) = -\frac{y(x)}{2x}\]
includes a cubic term of the dependent variable, making it nonlinear. These equations are known for modeling more complex phenomena in physics and biology but are also notoriously harder to solve.
Nonlinear equations do not generally have straightforward solutions and can exhibit behaviors such as sensitivity to initial conditions, making them challenging yet fascinating to study. They often require specific methods for finding solutions, such as perturbation techniques, numerical solutions, or, as in our exercise, the substitution method.
Substitution Method in Differential Equations
When a differential equation appears too complex to solve directly, one effective strategy is the substitution method. This process transforms the original equation into a new form that is hopefully easier to tackle.
As seen in our example, a substitution was made with
\[v(x) = y^2(x)\]
which simplifies the problem by reducing the complexity of the nonlinear terms. Following the substitution, derivative relationships, such as
\[v'(x) = 2y(x)y'(x)\]
are used to express the original equation in terms of the new variable.
The key to a successful substitution is to notice patterns or relationships in the equation that hint at a simpler underlying structure. This can turn an intractable equation into one that is more amenable to solution methods. However, choosing the correct substitution requires practice and familiarity with different types of differential equations.
As seen in our example, a substitution was made with
\[v(x) = y^2(x)\]
which simplifies the problem by reducing the complexity of the nonlinear terms. Following the substitution, derivative relationships, such as
\[v'(x) = 2y(x)y'(x)\]
are used to express the original equation in terms of the new variable.
The key to a successful substitution is to notice patterns or relationships in the equation that hint at a simpler underlying structure. This can turn an intractable equation into one that is more amenable to solution methods. However, choosing the correct substitution requires practice and familiarity with different types of differential equations.
Solving Differential Equations Step by Step
The process of solving differential equations step by step is vital for understanding the underlying behavior of the equation. Following a systematic approach allows students to build their skills incrementally and understand the reasoning behind each step.
The first step often involves rewriting the equation in a standard form, identifying whether the equation is linear, separable, or exact, which dictates the general solution approach. If the standard methods don't work, as with our exercise, more inventive techniques such as substitution are employed.
Each subsequent step delves deeper into the transformed equation, applying relevant mathematical concepts and simplifications until a solution is reached or the problem is deemed too complex for analytical methods. In those cases, numerical approximations may be necessary. Regardless of the method used, patience and attention to detail are crucial in solving differential equations step by step.
The first step often involves rewriting the equation in a standard form, identifying whether the equation is linear, separable, or exact, which dictates the general solution approach. If the standard methods don't work, as with our exercise, more inventive techniques such as substitution are employed.
Each subsequent step delves deeper into the transformed equation, applying relevant mathematical concepts and simplifications until a solution is reached or the problem is deemed too complex for analytical methods. In those cases, numerical approximations may be necessary. Regardless of the method used, patience and attention to detail are crucial in solving differential equations step by step.
Other exercises in this chapter
Problem 42
Solve the given differential equation. $$y^{\prime}+2 x^{-1} y=6 y^{2} x^{4}$$
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A second-order differential equation together with two auxiliary conditions imposed at different values of the independent variable is called a boundary- value
View solution Problem 44
Determine which of the five types of differential equations we have studied the given differential equation falls into, and use an appropriate technique to find
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The differential equation \(y^{\prime \prime}+y=0\) has the general solution \(y(x)=c_{1} \cos x+c_{2} \sin x\) (a) Show that the boundary-value problem \(y^{\p
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