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$$\frac{d y}{d x}=\frac{y(\ln y-\ln x+1)}{x}$$

Short Answer

Expert verified
The solution to the differential equation \(\frac{d y}{d x}=\frac{y(\ln y-\ln x+1)}{x}\) is given by \(y = x \cdot e^{(e^{-\ln(x) - C_1} - 1)}\), where \(C_1\) is the constant of integration.

Step by step solution

01

Identify the form of the differential equation

This is a separable differential equation of form \(\frac{dy}{dx} = f(x)g(y)\). We have \(\frac{dy}{dx} = \frac{y(\ln(y) - \ln(x) + 1)}{x}\). We can rewrite this as: \\(\frac{dy}{y(\ln(y) - \ln(x) + 1)} = \frac{dx}{x}\).
02

Integrate

Now we integrate both sides:\\(\int \frac{1}{y(1 + \ln(y) - \ln(x))} dy = \int \frac{1}{x} dx\). \Performing the integrations, we obtain \(- \ln(\ln(y)- \ln(x)+1) = \ln(x) + C_1\), where \(C_1\) is the constant of integration.
03

Rearrange to isolate y

Rearranging to isolate y, we take exponential of both sides: \\((\ln(y) - \ln(x) + 1) = e^{- \ln(x) - C_1}\). \Rearranging further, we get: \(\ln(y) = \ln(x) + e^{- \ln(x) - C_1}- 1 \). \Finally, we exponentiate to find the solution: \\(y = x \cdot e^{(e^{-\ln(x) - C_1} - 1)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration is a fundamental process in solving differential equations. Here, we integrate both sides of an equation to find a general solution. In this context, we're dealing with a separable differential equation, where integration is applied to expressions involving separated variables. For instance, once our original equation \(\frac{dy}{dx} = \frac{y(\ln(y) - \ln(x) + 1)}{x}\) is rewritten, both sides can be integrated separately.
  • Left Side: \(\int \frac{1}{y(1 + \ln(y) - \ln(x))} \,dy\)
  • Right Side: \(\int \frac{1}{x} \,dx\)
Performing these integrations may require different techniques. The integral on the right is straightforward, resulting in the natural logarithm \(\ln(x)\). The one on the left involves considering function inverses or substitution methods. This can be more complex, often necessitating clever algebra to simplify the fraction.
Through carefully performed integrations, we find \(- \ln(\ln(y) - \ln(x) + 1) = \ln(x) + C_1\), with \(C_1\) being an integration constant. This formula is integral to further solving for \(y\) in the equation, showcasing the power of integration techniques.
Differential Equation Solutions
A differential equation solution involves finding an equation describing a function, such as \(y=f(x)\), that satisfies the differential relationship. The equation given is separable, meaning it can be rearranged so that all terms involving \(y\) and \(dy\) are on one side and all terms involving \(x\) and \(dx\) are on the other. This property makes the integration process feasible and systematic.
The general solution for this separable differential equation was derived by first separating the variables:
  • \(\frac{dy}{y(\ln(y) - \ln(x) + 1)} = \frac{dx}{x}\)
Next, integrating both sides presented a solution containing a constant of integration, depicted logically as
\(- \ln(\ln(y) - \ln(x) + 1) = \ln(x) + C_1\).

When it comes to finding particular solutions, initial conditions might be used. This constant \(C_1\) would then be resolved to ensure the solution meets specific conditions. These solutions are invaluable in fields like physics and engineering, allowing for modeling and prediction of real-world systems.
Isolation of Variables
Isolation of variables is an essential step in solving differential equations, especially separable ones. This process involves rearranging the equation so that each type of variable is grouped together, typically making one side dependent on \(x\) and the other on \(y\). For the given differential equation \(\frac{d y}{d x}=\frac{y(\ln y-\ln x+1)}{x}\), we accomplish this in Step 1: \(\frac{dy}{y(\ln(y) - \ln(x) + 1)} = \frac{dx}{x}\).

Isolating variables allows each side of the equation to be integrated separately. This clarity is why isolating variables is a fundamental technique when dealing with differential equations. After the integration, the challenge is often to rearrange the result to solve for the dependent variable \(y\).
Through exponential techniques and some algebraic manipulation, we arrive at the final form of the solution:
  • From \(\ln(y) = \ln(x) + e^{- \ln(x) - C_1} - 1\) to \(y = x \cdot e^{(e^{-\ln(x) - C_1} - 1)}\)
This final step ensures that the function \(y\) is explicitly expressed in terms of \(x\). The isolation and manipulation techniques alleviate potential complexity, making the solution more accessible. It exemplifies how solving differential equations is both analytical and methodical.

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Most popular questions from this chapter

In Problems \(9-20\) , determine whether the equation is exact. If it is, then solve it. $$\left(2 x+\frac{y}{1+x^{2} y^{2}}\right) d x+\left(\frac{x}{1+x^{2} y^{2}}-2 y\right) d y=0$$

Riccati Equation. An equation of the form $$\frac{d y}{d x}=P(x) y^{2}+Q(x) y+R(x)$$ is called a generalized Riccati equation. (a) If one solution_say,$$u(x)-\text { of }(18)$$ is known, show that the substitution $$y=u+1 / v \text { reduces }(18)$$ to a linear equation in v. (b) Given that $$u(x)=x$$ is a solution to $$\frac{d y}{d x}=x^{3}(y-x)^{2}+\frac{y}{x}$$ use the result of part (a) to find all the other solutions to this equation. (The particular solution $$u(x)=x$$ can be found by inspection or by using a Taylor series method; see Section 8.1.)

Compound Interest. If \(P(t)\) is the amount of dollars in a savings bank account that pays a yearly interest rate of \(r \%\) compounded continuously, then $$\frac{d P}{d t}=\frac{r}{100} P, \quad t$$ in years. Assume the interest is 5\(\%\) annually, \(P(0)=\$ 1000,\) and no monies are withdrawn. (a) How much will be in the account after 2 yr? (b) When will the account account every 12 months, (c) If \(\$ 1000\) is added to the account every 12 months, how much will be in the account after 3\(\frac{1}{2}\) yr?

Uniqueness Questions. In Chapter 1 we indicated that in applications most initial value problems will have a unique solution. In fact, the existence of unique solutions was so important that we stated an existence and uniqueness theorem, Theorem 1, page 11. The method for separable equations can give us a solution, but it may not give us all the solutions (also see Problem 30). To illustrate this, consider the equation \( d y / d x=y^{1 / 3} \). (a) Use the method of separation of variables to show that $$ y=\left(\frac{2 x}{3}+C\right)^{3 / 2} $$ is a solution. (b) Show that the initial value problem \( d y / d x \) = \( y^{1 / 3} \) with \( y(0)=0 \) is satisfied for \( C=0 \) by \( y=(2 x / 3)^{3 / 2} \) for \( x \geq 0 \). (c) Now show that the constant function \( y \equiv 0 \) also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution. (d) Finally, show that the conditions of Theorem 1 on page 11 are not satisfied. (The solution \( y \equiv 0 \) was lost because of the division by zero in the separation process.)

In Problems \(1-6,\) identify the equation as separable, linear, exact, or having an integrating factor that is a function of either \(x\) alone or \(y\) alone. $$\left(y^{2}+2 x y\right) d x-x^{2} d y=0$$

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