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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

Newton's Law of Cooling. According to Newton's law of cooling, if an object at temperature \( T \) is immersed in a medium having the constant temperature \( M \), then the rate of change of \( T \) is proportional to the difference of temperature \( M-T \). This gives the differential equation $$ d T / d t=k(M-T) $$ (a) Solve the differential equation for \( T \) (b) A thermometer reading \( 100^{\circ} \mathrm{F} \) is placed in a medium having a constant temperature of \( 70^{\circ} \mathrm{F} \). After 6 min, the thermometer reads \( 80^{\circ} \mathrm{F} \). What is the reading after 20 min? (Further applications of Newton's law of cooling appear in Section 3.3.)

As discussed in calculus, certain indefinite integrals (antiderivatives) such as \( \int e^{x^{x}} d x \) cannot be expressed in finite terms using elementary functions. When such an integral is encountered while solving a differential equation, it is often helpful to use definite integration (integrals with variable upper limit). For example, consider the initial value problem $$ \frac{d y}{d x}=e^{x^{2}} y^{2}, \quad y(2)=1 $$ The differential equation separates if we divide by \( y^{2} \) and multiply by \( d x \). We integrate the separated equation from \( x=2 \) to \( x=x_{1} \) and find $$ \int_{x=2}^{x=x_{1}} e^{x^{2}} d x=\int_{x=2}^{x=x_{1}} \frac{d y}{y^{2}} $$ $$ =-\frac{1}{y} \left| \begin{array}{l}{x=x_{1}} \\ {x=2}\end{array}\right. $$ $$ =-\frac{1}{y\left(x_{1}\right)}+\frac{1}{y(2)} $$ If we let \( t \) be the variable of integration and replace \( x_{1} \) by \( x \) and \( y(2) \) by 1, then we can express the solution to the initial value problem by $$ y(x)=\left(1-\int_{2}^{x} e^{t^{2}} d t\right)^{-1} $$ Use definite integration to find an explicit solution to the initial value problems in parts (a)-(c). (a) \( d y / d x=e^{x^{2}}, \quad y(0)=0 \) (b) \( d y / d x=e^{x^{2}} y^{-2}, \quad y(0)=1 \) (c) \( d y / d x=\sqrt{1+\sin x}\left(1+y^{2}\right), \quad y(0)=1 \) (d) Use a numerical integration algorithm (such as Simpson's rule, described in Appendix C) to approximate the solution to part (b) at \( x=0.5 \) to three decimal places.

In Problems \(1-8,\) classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification. $$\left(y e^{x y}+2 x\right) d x+\left(x e^{x y}-2 y\right) d y=0$$

Consider the initial value problem $$ \frac{d y}{d x}+\sqrt{1+\sin ^{2} x} y=x, \quad y(0)=2 $$. (a) Using definite integration, show that the integrating factor for the differential equation can be written as $$ \mu(x)=\exp \left(\int_{0}^{x} \sqrt{1+\sin ^{2} t} d t\right) $$ and that the solution to the initial value problem is $$ y(x)=\frac{1}{\mu(x)} \int_{0}^{x} \mu(s) s d s+\frac{2}{\mu(x)} $$. (b) Obtain an approximation to the solution at \( x=1 \) by using numerical integration (such as Simpson's rule, Appendix C) in a nested loop to estimate values of \( \mu(x) \) and, thereby, the value of $$ \int_{0}^{1} \mu(s) s d s $$. [Hint: First, use Simpson's rule to approximate \( \mu(x) \) at \( x=0.1,0.2, \dots, 1 \). Then use these values and apply Simpson's rule again to approximate \( \int_{0}^{1} \mu(s) s d s . ] \) (c) Use Euler's method (Section 1.4) to approximate the solution at \( x=1 \), with step sizes \( h=0.1 \) and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]

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.)

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