y'=2x+y. Find a solution whose graph is also an isocline.
To find a solution whose graph is also an isocline, we need to determine the equation of the isocline and then find a solution that satisfies that equation.
An isocline represents a set of points where the slope of the solution curve is constant. In this case, we want the slope of the solution curve to be equal to 2x+y.
For an isocline, the derivative of the solution function with respect to x must be equal to the given slope. So, we want to find a solution that satisfies the condition y' = 2x + y.
Let's solve this differential equation:
Separate the variables:
dy/dx - y = 2x
This is a first-order linear nonhomogeneous differential equation. To solve it, we'll use an integrating factor. Let's multiply both sides of the equation by e^(-x):
e^(-x) * dy/dx - e^(-x) * y = 2x * e^(-x)
Now, the left side can be written as the derivative of (e^(-x) * y):
d/dx [e^(-x) * y] = 2x * e^(-x)
Integrating both sides with respect to x, we get:
e^(-x) * y = ∫(2x * e^(-x)) dx
Now, let's find the integral of (2x * e^(-x)):
∫(2x * e^(-x)) dx = -2(x + 1)e^(-x) + C
where C is the constant of integration.
Substituting this back into the equation:
e^(-x) * y = -2(x + 1)e^(-x) + C
To get y alone, we divide both sides by e^(-x):
y = -2(x + 1) + Ce^x
So the solution to the differential equation y' = 2x + y, whose graph is also an isocline, is given by y = -2(x + 1) + Ce^x, where C is an arbitrary constant.