Find the general solution of the equation [tex]\frac{d y}{d t}=\frac{y-4 t}{t-y}[/tex]

Substitute v = t y ​ , so y = v t and d t d y ​ = v + t d t d v ​ .
Separate variables and integrate both sides using partial fractions.
Substitute back v = t y ​ .
The general solution is ( y − 2 t ) ( y + 2 t ) 3 = C ​ .

Explanation

Problem Analysis The given differential equation is d t d y ​ = t − y y − 4 t ​ . This is a first-order homogeneous differential equation.

Substitution To solve this, we can rewrite the equation as d t d y ​ = 1 − t y ​ t y ​ − 4 ​ . Now, let's use the substitution v = t y ​ , which means y = v t . Differentiating with respect to t , we get d t d y ​ = v + t d t d v ​ .

Transformed Equation Substituting these into the differential equation, we have v + t d t d v ​ = 1 − v v − 4 ​ .

Separation of Variables Now, let's separate the variables: t d t d v ​ = 1 − v v − 4 ​ − v = 1 − v v − 4 − v ( 1 − v ) ​ = 1 − v v − 4 − v + v 2 ​ = 1 − v v 2 − 4 ​ .

Partial Fractions So, we have v 2 − 4 1 − v ​ d v = t 1 ​ d t . Now, we integrate both sides. To integrate the left side, we use partial fractions: v 2 − 4 1 − v ​ = ( v − 2 ) ( v + 2 ) 1 − v ​ = v − 2 A ​ + v + 2 B ​ . Then 1 − v = A ( v + 2 ) + B ( v − 2 ) .

Solving for Coefficients To find A and B , we can use the following: If v = 2 , then 1 − 2 = A ( 2 + 2 ) , so − 1 = 4 A and A = − 4 1 ​ . If v = − 2 , then 1 − ( − 2 ) = B ( − 2 − 2 ) , so 3 = − 4 B and B = − 4 3 ​ .

Integration Thus, ∫ v 2 − 4 1 − v ​ d v = ∫ v − 2 − 4 1 ​ ​ + v + 2 − 4 3 ​ ​ d v = − 4 1 ​ ln ∣ v − 2∣ − 4 3 ​ ln ∣ v + 2∣ . Also, ∫ t 1 ​ d t = ln ∣ t ∣ + C .

Simplification So, we have − 4 1 ​ ln ∣ v − 2∣ − 4 3 ​ ln ∣ v + 2∣ = ln ∣ t ∣ + C . Multiplying by − 4 , we get ln ∣ v − 2∣ + 3 ln ∣ v + 2∣ = − 4 ln ∣ t ∣ + C 1 ​ , where C 1 ​ = − 4 C .

Exponentiation Then ln ∣ v − 2∣ + ln ∣ v + 2 ∣ 3 = ln ∣ t ∣ − 4 + C 1 ​ , so ln ∣ ( v − 2 ) ( v + 2 ) 3 ∣ = ln ∣ t ∣ − 4 + C 1 ​ . Exponentiating, we get ∣ ( v − 2 ) ( v + 2 ) 3 ∣ = e l n ∣ t ∣ − 4 + C 1 ​ = e C 1 ​ ∣ t ∣ − 4 = C 2 ​ ∣ t ∣ − 4 , where C 2 ​ = e C 1 ​ .

Back-Substitution Then ( v − 2 ) ( v + 2 ) 3 = C 3 ​ t − 4 , where C 3 ​ = ± C 2 ​ . Substituting v = t y ​ , we get ( t y ​ − 2 ) ( t y ​ + 2 ) 3 = C 3 ​ t − 4 . Multiplying by t 4 , we get ( t y ​ − 2 ) ( t y ​ + 2 ) 3 t 4 = ( y − 2 t ) ( y + 2 t ) 3 = C 3 ​ .

Final Solution So, the general solution is ( y − 2 t ) ( y + 2 t ) 3 = C , where C is an arbitrary constant.


Examples
In physics, this type of differential equation can model the motion of a particle in a central force field where the force depends on the distance from the center. The solution helps describe the particle's trajectory over time, which is crucial in understanding celestial mechanics or atomic behavior. For example, understanding the motion of satellites around a planet or electrons around an atom involves solving similar differential equations.

Answered by GinnyAnswer | 2025-07-08

The general solution to the differential equation d t d y ​ = t − y y − 4 t ​ is given by ( y − 2 t ) ( y + 2 t ) 3 = C , where C is an arbitrary constant. This is derived using a substitution method and separation of variables. The process involves rewriting the equation, substituting variables, and integrating both sides.
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Answered by Anonymous | 2025-08-07