Solve $y'(x) = a + b \cdot y(x) \cdot x$. When $x= 0$ : $y = 0$. And $a> 0$, $b > 0$
The solution to the differential equation \( y'(x) = a + bx \cdot y(x) \) with the initial condition \( y(0) = 0 \) and given \( a > 0, b > 0 \) is:
$ y(x) = \dfrac{\sqrt{2} \sqrt{\pi} a e^{\dfrac{b x^2}{2}} \text{erf}\left(\dfrac{\sqrt{2} \sqrt{b} x}{2}\right)}{2 \sqrt{b}} $
Here, \(\text{erf}\) denotes the error function.
Full solution:
Certainly! To solve the differential equation \( y'(x) = a + bx \cdot y(x) \) with the initial condition \( y(0) = 0 \), and given \( a > 0, b > 0 \), we can follow these steps:
1. **Setting up the Equation:**
The given differential equation is a first-order linear ordinary differential equation:
$ y'(x) = a + bx \cdot y(x) $
2. **Rewriting in Standard Form:**
The standard form of a linear first-order differential equation is \( y' + P(x)y = Q(x) \). Here, we can rearrange our equation to match this form:
$ y'(x) - bx \cdot y(x) = a $
3. **Finding an Integrating Factor:**
The integrating factor for an equation of this form is \( e^{\int -P(x) dx} \). In our case, \( P(x) = -bx \), so the integrating factor is:
$ \mu(x) = e^{\int bx dx} = e^{\dfrac{b x^2}{2}} $
4. **Multiplying Through by the Integrating Factor:**
We multiply every term in the standard form equation by the integrating factor:
$ e^{\dfrac{b x^2}{2}} y' - bxe^{\dfrac{b x^2}{2}} y = ae^{\dfrac{b x^2}{2}} $
5. **Recognizing the Left Side as a Derivative:**
The left-hand side of our equation is now the derivative of the product \( \mu(x)y(x) \):
$ \dfrac{d}{dx}\left( e^{\dfrac{b x^2}{2}} y \right) = ae^{\dfrac{b x^2}{2}} $
6. **Integrating Both Sides:**
We integrate both sides with respect to \( x \):
$ \int \dfrac{d}{dx}\left( e^{\dfrac{b x^2}{2}} y \right) dx = \int ae^{\dfrac{b x^2}{2}} dx $
$ e^{\dfrac{b x^2}{2}} y = \int ae^{\dfrac{b x^2}{2}} dx $
7. **Solving the Integral:**
The integral on the right side involves an exponential function of a quadratic, which is related to the error function \( \text{erf}(x) \):
$ \int ae^{\dfrac{b x^2}{2}} dx = \dfrac{\sqrt{2} \sqrt{\pi} a \text{erf}\left(\dfrac{\sqrt{2} \sqrt{b} x}{2}\right)}{2 \sqrt{b}} + C $
8. **Solving for \( y(x) \):**
Rearranging for \( y(x) \), we get:
$ y(x) = \dfrac{\sqrt{2} \sqrt{\pi} a \text{erf}\left(\dfrac{\sqrt{2} \sqrt{b} x}{2}\right)}{2 \sqrt{b} e^{\dfrac{b x^2}{2}}} + \dfrac{C}{e^{\dfrac{b x^2}{2}}} $
9. **Applying the Initial Condition \( y(0) = 0 \):**
When \( x = 0 \), the error function \(\text{erf}(0)\) is zero, simplifying our equation. Applying the initial condition, we find:
$ 0 = \dfrac{C}{e^0} $
$ C = 0 $
10. **Final Solution:**
With \( C = 0 \), the solution is:
$ y(x) = \dfrac{\sqrt{2} \sqrt{\pi} a \text{erf}\left(\dfrac{\sqrt{2} \sqrt{b} x}{2}\right)}{2 \sqrt{b} e^{\dfrac{b x^2}{2}}} $
This is the complete solution to the given differential equation with the specified initial condition.
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