Is dy/dt +(t*(y^2)) = 0 a non-linear differntial equation?
Is it because of the y^2 term involved with t?
Many thanks!
yes
just review the topic -- google is your friend
dy / dt + t ∙ y² = 0
Substitute dy / dt with y′
y′ + t ∙ y² = 0
Subtract t ∙ y² to both sides
y′ + t ∙ y² - t ∙ y² = 0 - t ∙ y²
y′ = - t ∙ y²
Divide both sides by y²
y′ / y² = - t
( 1 / y² ) ∙ y′ = - t
This mean:
dy / dt + t ∙ y² = 0 is the same as ( 1 / y² ) ∙ y′ = - t
A first order separable Ordinary Differential Equation has the form of:
N(y) · y′ = M(x)
In this case:
N(y) · y′ = M(t)
where:
N(y) = 1 / y² , M(t) = - t
1 / y² is nonlinear function
So dy / dt + t ∙ y² = 0 is first order nonlinear Ordinary Differential Equation.
Thank you very much @Bosnian and oobleck
Yes, the given differential equation dy/dt + t*(y^2) = 0 is a non-linear differential equation.
To understand why, let's break down the equation. A differential equation is linear if it can be expressed as a linear combination of the unknown function y and its derivatives, where the coefficients can only be functions of the independent variable t.
In this case, the term t*(y^2) is the non-linear part of the equation. The presence of the y^2 term makes it non-linear because the power of y is greater than 1. In a linear differential equation, the unknown function y and its derivatives are raised to only the power of 1 or 0.
Therefore, the given differential equation is non-linear due to the y^2 term involved with t.