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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!
3 answers
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.
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