The solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is
x-1
The problem asks us to find the specific solution of the given differential equation that passes through a particular point. This is an initial value problem, where we first find the general solution and then use the given point to determine the constant of integration.
Differential Equation Analysis
We are given the differential equation:
\[{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0\]
This is a first-order linear differential equation, but it can also be solved using the method of separation of variables because it is homogeneous. Our goal is to isolate \({\rm{y}}\) terms on one side and \({\rm{x}}\) terms on the other.
Separating Variables
First, let's rearrange the equation to separate the variables. We move the \({\rm{y}}\) term to the right side:
\[{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = -{\rm{y}}\]
Now, we want all \({\rm{y}}\) terms with \({\rm{dy}}\) and all \({\rm{x}}\) terms with \({\rm{dx}}\). We can divide both sides by \({\rm{y}}\) and by \({\rm{x}}\), and multiply by \({\rm{dx}}\):
\[\frac{{{\rm{dy}}}}{{{\rm{y}}}} = -\frac{{{\rm{dx}}}}{{{\rm{x}}}}\]
This form is suitable for integration.
Integrating Both Sides
Next, we integrate both sides of the separated equation:
\[\int \frac{{{\rm{dy}}}}{{{\rm{y}}}} = \int -\frac{{{\rm{dx}}}}{{{\rm{x}}}}\]
Performing the integration, we get:
\[\ln|{\rm{y}}| = -\ln|{\rm{x}}| + {\rm{C}}\]
where \({\rm{C}}\) is the constant of integration. We can simplify the right-hand side using logarithm properties: \(-\ln|{\rm{x}}| = \ln|{\rm{x}}|^{{\rm{-1}}}\). So the equation becomes:
\[\ln|{\rm{y}}| = \ln|{\rm{x}}|^{{\rm{-1}}} + {\rm{C}}\]
To combine the logarithmic terms, let's express the constant \({\rm{C}}\) as \(\ln|{\rm{K}}|\), where \({\rm{K}}\) is another arbitrary positive constant:
\[\ln|{\rm{y}}| = \ln|{\rm{x}}|^{{\rm{-1}}} + \ln|{\rm{K}}|\)]
Using the logarithm property \(\ln({\rm{a}}) + \ln({\rm{b}}) = \ln({\rm{ab}})\):
\[\ln|{\rm{y}}| = \ln\left(|{\rm{K}}| \cdot |{\rm{x}}|^{{\rm{-1}}}\right)\]
Exponentiating both sides (taking \({\rm{e}}\) to the power of both sides) to remove the logarithm:
\[{\rm{y}} = {\rm{K}} \cdot {\rm{x}}^{{\rm{-1}}}\]
Which can also be written as:
\[{\rm{y}} = \frac{{\rm{K}}}{{\rm{x}}}\]
or
\[{\rm{xy}} = {\rm{K}}\]
This is the general solution of the differential equation.
Applying Initial Condition
We are given that the solution passes through the point (1,1). This means when \({\rm{x}} = 1\), \({\rm{y}} = 1\). We use this information to find the specific value of the constant \({\rm{K}}\).
Substitute \({\rm{x}} = 1\) and \({\rm{y}} = 1\) into the general solution \({\rm{xy}} = {\rm{K}}\):
\[(1)(1) = {\rm{K}}\]
\[{\rm{K}} = 1\]
Final Solution
Now that we have the value of \({\rm{K}}\), we substitute it back into the general solution \({\rm{y}} = \frac{{\rm{K}}}{{\rm{x}}}\):
\[{\rm{y}} = \frac{1}{{\rm{x}}}\]
This can also be expressed using a negative exponent:
\[{\rm{y}} = {\rm{x}}^{{\rm{-1}}}\]
This matches one of the given options.
The final answer is x-1.
Important Questions from Order and Degree of a Differential Equation
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a
The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin x\left( {\frac{{dy}}{{dx}}} \right) + y = 0\) is:
In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ
\(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
The above equation isThe order and degree of the differential equation
\(\frac{{{d^3}y}}{{d{x^3}}} + 4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}= 0\;\)
