Problem 60
Question
Consider the differential equation $$ \frac{d y}{d x}=\frac{x+2 y-1}{2 x-y+3} $$ (a) Show that the change of variables defined by $$ x=u-1, \quad y=v+1 $$ transforms Equation (1.8.20) into the homogeneous equation $$ \frac{d v}{d u}=\frac{u+2 v}{2 u-v} $$ (b) Find the general solution to Equation and hence, solve Equation ( 1.8 .20 ).
Step-by-Step Solution
Verified Answer
The general solution for the given differential equation after applying the change of variables and transforming it into a homogenous equation is given by:
$$
\frac{x+2y}{1} = k(2x-y+1)
$$
where \(k\) is an arbitrary constant.
1Step 1: Apply the change of variables
To transform the given non-homogeneous equation into a homogeneous equation, let's perform the change of variables as suggested:
$$
x = u - 1, \quad y = v + 1
$$
We need to find the new relation between \(dy/dx\) and \(dv/du\). We can differentiate the second equation with respect to \(u\) to find \(dv/du\):
$$
\frac{dy}{du} = \frac{dv}{du}
$$
As we have \(dy/dx\) and \(dy/du\), we can compute \(dx/du\) and make the substitution:
$$
\frac{dx}{du} = 1 \Rightarrow dx = du
$$
Now, find the expression for \(\frac{dy}{dx}\) in terms of \(\frac{dv}{du}\):
$$
\frac{dy}{dx}=\frac{dv/du}{dx/du}=\frac{dv}{du}
$$
2Step 2: Transform the differential equation
Now let's substitute the values of \(x\), \(y\), and \(\frac{dy}{dx}\) into the given differential equation:
$$
\frac{dv}{du} = \frac{(u-1)+2(v+1)-1}{2(u-1)-(v+1)+3}
$$
Simplify the equation:
$$
\frac{dv}{du} = \frac{u+2v}{2u-v}
$$
As we can see, the resulting equation is homogeneous.
3Step 3: Solve the homogeneous equation
To solve the homogeneous equation, we can reformulate it as a separable equation:
$$
\frac{dv}{u+2v} = \frac{du}{2u-v}
$$
Now we can integrate both sides of the equation:
$$
\int\frac{dv}{u+2v} = \int\frac{du}{2u-v}
$$
To solve these integrals, we can use the substitution method. For the left side, let \(w = u+2v\) and \(dw = 2dv\):
$$
\frac{1}{2}\int\frac{dw}{w} = \int\frac{2u-du}{2u(2u-v)}
$$
For the right side, let \(z = 2u-v\) and \(dz = 2du-dv\):
$$
\frac{1}{2}\int\frac{dw}{w} = \int\frac{dz}{2u(2u+z)}
$$
Integrate both sides:
$$
\frac{1}{2}\ln|w| = \frac{1}{2}\ln|z| + C
$$
4Step 4: Solve for the constant and find the general solution
Now, substitute the original expressions for \(w\) and \(z\) back:
$$
\frac{1}{2}\ln|u+2v| = \frac{1}{2}\ln|2u-v| + C
$$
We can rearrange this equation as follows to find the general solution:
$$
\frac{u+2v}{1} = k(2u-v)
$$
where \(k = e^C\) is the integration constant.
5Step 5: Transform back to the original variables
Finally, we need to transform back to the original variables to obtain the general solution for the original differential equation. Substitute the expressions \(x=u-1\) and \(y=v+1\) back to the equation:
$$
\frac{(u-1)+2(v+1)}{1} = k(2(u-1)-v)
$$
Simplify the equation to get the final solution:
$$
\frac{x+2y}{1} = k(2x-y+1)
$$
Key Concepts
Change of VariablesHomogeneous EquationsGeneral SolutionSeparable Equations
Change of Variables
A change of variables is a helpful technique in solving complex differential equations. By substituting different variables into the equation, we may simplify its form.
In the original exercise, we use:
The change of variables here allows us to recognize patterns and structures within the differential equation, making it easier to identify solutions. It is like swapping a tough puzzle piece for one that reveals the bigger picture.
In the original exercise, we use:
- \(x = u - 1\)
- \(y = v + 1\)
The change of variables here allows us to recognize patterns and structures within the differential equation, making it easier to identify solutions. It is like swapping a tough puzzle piece for one that reveals the bigger picture.
Homogeneous Equations
Homogeneous equations play a vital role in differential equations, representing cases where every term is of the same degree.
In our problem, after applying the change of variables, the equation turns into:\[\frac{d v}{d u} = \frac{u + 2v}{2u - v}\]This is a homogeneous equation because each term is a combination of \(u\) and \(v\) with equal degree weightage.
This transformation is essential as homogeneous equations can often be solved more straightforwardly or with known methods. Homogeneity implies symmetry which helps in simplifying expressions and often allows the equation to be solved using special techniques, like separation of variables here.
In our problem, after applying the change of variables, the equation turns into:\[\frac{d v}{d u} = \frac{u + 2v}{2u - v}\]This is a homogeneous equation because each term is a combination of \(u\) and \(v\) with equal degree weightage.
This transformation is essential as homogeneous equations can often be solved more straightforwardly or with known methods. Homogeneity implies symmetry which helps in simplifying expressions and often allows the equation to be solved using special techniques, like separation of variables here.
General Solution
The general solution of a differential equation gives us a family of potential solutions, expressed with arbitrary constants. After transforming our equation into the homogeneous form, the general solution becomes easier to find.
By integrating:\[\int\frac{dv}{u+2v} = \int\frac{du}{2u-v}\]we solve these integrals to reach:\[\frac{1}{2}\ln|u+2v| = \frac{1}{2}\ln|2u-v| + C\]By solving for the constant and translating back to our original variables, we end up with:\[\frac{x+2y}{1} = k(2x-y+1)\]where \(k = e^C\) is the constant. This solution encompasses all possible specific solutions depending on initial conditions or additional constraints.
By integrating:\[\int\frac{dv}{u+2v} = \int\frac{du}{2u-v}\]we solve these integrals to reach:\[\frac{1}{2}\ln|u+2v| = \frac{1}{2}\ln|2u-v| + C\]By solving for the constant and translating back to our original variables, we end up with:\[\frac{x+2y}{1} = k(2x-y+1)\]where \(k = e^C\) is the constant. This solution encompasses all possible specific solutions depending on initial conditions or additional constraints.
Separable Equations
Separable equations are a type of differential equation where variables can be separated on either side of the equation. This property makes them especially easy to solve, as integration can be applied directly.
Our homogeneous equation can be written in separable form:\[\frac{dv}{u+2v} = \frac{du}{2u-v}\]This allows each side to be integrated independently. When variables are separable, the integration becomes more straightforward since it reduces a complex equation into manageable parts.
The separation of variables into a form that separates \(u\) and \(v\) greatly facilitates finding a solution, as seen in this example, where it leads directly to integration and the general solution derivation.
Our homogeneous equation can be written in separable form:\[\frac{dv}{u+2v} = \frac{du}{2u-v}\]This allows each side to be integrated independently. When variables are separable, the integration becomes more straightforward since it reduces a complex equation into manageable parts.
The separation of variables into a form that separates \(u\) and \(v\) greatly facilitates finding a solution, as seen in this example, where it leads directly to integration and the general solution derivation.
Other exercises in this chapter
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