What is the solution of the differential equation \(\rm\ln \left(\dfrac{dy}{dx}\right) =x ?\)
y = e x+ c
Solving the Differential Equation \( \ln \left(\frac{dy}{dx}\right) = x \)
The question asks us to find the solution to the given differential equation: \(\ln \left(\dfrac{dy}{dx}\right) =x\). This is a first-order differential equation.
To solve this differential equation, we first need to isolate the derivative term, \(\dfrac{dy}{dx}\). The equation involves a natural logarithm of the derivative. We can remove the logarithm by exponentiating both sides of the equation with base \(e\). Recall that \(e^{\ln A} = A\) for any positive A.
Given: \(\ln \left(\dfrac{dy}{dx}\right) =x\)
Exponentiate both sides with base \(e\):
\(e^{\ln \left(\frac{dy}{dx}\right)} = e^x\)
This simplifies to:
\(\dfrac{dy}{dx} = e^x\)
Now we have a simple first-order differential equation where the derivative is expressed as a function of \(x\). We can solve this by separating variables and integrating.
Rewrite the equation by treating \(dy\) and \(dx\) as differentials:
\(dy = e^x \, dx\)
Now, integrate both sides of the equation:
\(\int dy = \int e^x \, dx\)
Integrating the left side with respect to \(y\) gives \(y\). Integrating the right side with respect to \(x\) gives \(e^x\). Remember to include the constant of integration, commonly denoted by \(c\) or \(C\), on one side of the equation.
\(y = e^x + c\)
This equation \(y = e^x + c\) represents the general solution to the given differential equation \(\ln \left(\dfrac{dy}{dx}\right) =x\). The constant \(c\) can be any real number, determining a family of solutions.
Now let's compare our derived solution with the given options:
- Option 1: \(y = e^x + c\)
- Option 2: \(y = e^{-x} + c\)
- Option 3: \(y = \ln x + c\)
- Option 4: \(y = 2 \ln x + c\)
Our derived solution \(y = e^x + c\) exactly matches Option 1.
Revision Table: Key Steps in Solving ODEs
| Step | Description | Applied to \(\ln \left(\frac{dy}{dx}\right) = x\) |
|---|---|---|
| Identify ODE Type | Determine if it's first-order, second-order, separable, linear, etc. | First-order, can be made separable. |
| Isolate Derivative | Manipulate the equation to get \(\frac{dy}{dx}\) by itself. | \(e^{\ln \left(\frac{dy}{dx}\right)} = e^x \implies \frac{dy}{dx} = e^x\). |
| Separate Variables | Rearrange terms to have all \(y\) terms with \(dy\) and all \(x\) terms with \(dx\). | \(dy = e^x \, dx\). |
| Integrate Both Sides | Apply the integral operator to both sides of the separated equation. | \(\int dy = \int e^x \, dx\). |
| Solve Integrals | Perform the integration for both sides. | \(y = e^x + c\). |
| Include Constant | Always add a constant of integration (\(c\) or \(C\)) when solving indefinite integrals. | Added \(+ c\) to the right side. |
Additional Information: Separable Differential Equations
A first-order differential equation of the form \(\dfrac{dy}{dx} = f(x, y)\) is called separable if the function \(f(x, y)\) can be written as a product of a function of \(x\) only and a function of \(y\) only. That is, \(f(x, y) = g(x)h(y)\).
In our case, after exponentiating, we got \(\dfrac{dy}{dx} = e^x\). Here, \(g(x) = e^x\) and \(h(y) = 1\). Since it can be written in the form \(g(x)h(y)\), it is a separable differential equation.
The general method for solving separable differential equations \(\dfrac{dy}{dx} = g(x)h(y)\) is:
- Separate variables: \(\dfrac{dy}{h(y)} = g(x) \, dx\) (provided \(h(y) \neq 0\)).
- Integrate both sides: \(\int \dfrac{dy}{h(y)} = \int g(x) \, dx\).
- Solve the integrals to find the relationship between \(y\) and \(x\), including the constant of integration.
For \(\dfrac{dy}{dx} = e^x\), \(h(y) = 1\), so the separation is \(dy = e^x \, dx\), which is what we performed.
The integration of \(e^x\) is a fundamental result in calculus:
\(\int e^x \, dx = e^x + c\)
This property makes solving differential equations involving \(e^x\) often straightforward once the derivative is isolated.
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