Find the solution t the differential equation
dz/dt = 2te^z that passes through the origin.
My answer is -In (1-t^2), but my answer is populating as incorrect. Can someone help please?
dz/dt = 2t e^z
e^-z dz = 2t dt
-e^-z = t^2 + c
e^-z = c-t^2
-z = ln(c-t^2)
z = -ln(c-t^2)
at(0,0) 0 = -ln(c)
c = 1
z = -ln(1-t^2)
I like your answer. Maybe they want
z = ln(1/(1-t^2))
Well, it seems like your answer isn't quite right, but that's okay! Differential equations can be tricky sometimes. Let's work through it together and see if we can find the correct solution.
The given differential equation is dz/dt = 2te^z. To solve this, let's separate the variables:
dz/e^z = 2t dt
Now, let's integrate both sides:
∫dz/e^z = 2∫t dt
To integrate the left side, we can use the substitution u = e^z:
du = e^z dz
So, the left side becomes:
∫du = ∫2t dt
Integrating both sides, we get:
u = t^2 + C1
Now, substituting back in u = e^z, we have:
e^z = t^2 + C1
To find the particular solution that passes through the origin, we can substitute t = 0 and z = 0 into the equation:
e^0 = 0^2 + C1
1 = C1
So, the particular solution that passes through the origin is:
e^z = t^2 + 1
Taking the natural logarithm of both sides, we get:
z = ln(t^2 + 1)
And there you have it! The correct solution to the differential equation dz/dt = 2te^z that passes through the origin is z = ln(t^2 + 1). Keep in mind that there might be some small calculation errors, so double-check your work just to be sure.
To find the solution to the given differential equation, we can use separation of variables.
Starting with the original equation: dz/dt = 2te^z
Let's separate the variables by moving all terms involving z to one side and all terms involving t to the other side:
e^(-z)dz = 2tdt
Now, we can integrate both sides:
∫e^(-z)dz = ∫2tdt
To integrate the left side, we can use the substitution u = -z, which gives us du = -dz:
-∫e^u du = ∫2tdt
- e^u = t^2 + C1, where C1 is the constant of integration.
Substituting back u = -z:
- e^(-z) = t^2 + C1
To solve for z, we take the natural logarithm of both sides:
ln(- e^(-z)) = ln(t^2 + C1)
Simplifying the left side:
-ln(e^(-z)) = ln(t^2 + C1)
z = -ln(t^2 + C1)
Now, we need to use the initial condition that the solution passes through the origin, which means z(0) = 0:
0 = -ln((0)^2 + C1)
0 = -ln(C1)
C1 = 1
Substituting this back into the equation for z:
z = -ln(t^2 + 1)
So, the solution to the given differential equation that passes through the origin is z = -ln(t^2 + 1).
To solve the differential equation dz/dt = 2te^z, we can use separation of variables.
First, we can rewrite the equation as:
e^(-z)dz = 2t dt
Next, we integrate both sides with respect to their respective variables:
∫ e^(-z)dz = ∫ 2t dt
To integrate the left side, we can use the substitution u = -z, which means du = -dz. This simplifies the left side to:
-∫ e^u du = -e^u + C1
Where C1 is the constant of integration.
For the right side, we integrate 2t with respect to t:
∫ 2t dt = t^2 + C2
Where C2 is another constant of integration.
Putting it all together, we have:
-e^(-z) = t^2 + C
where C = C2 - C1.
Now, we need to solve for z. Multiply both sides by -1 to get rid of the negative sign:
e^(-z) = -t^2 - C
Take the natural logarithm of both sides:
-ln(e^(-z)) = ln(-t^2 - C)
-simplify-
z = ln(1/(-t^2 - C))
Now, we are given that the solution passes through the origin, which means z(0) = 0. Let's substitute these values into our equation to solve for C:
0 = ln(1/(-0^2 - C))
0 = ln(1/C)
Since ln(1/C) = 0, C = 1.
Now, we can substitute C = 1 back into the equation:
z = ln(1/(-t^2 - 1))
Therefore, the solution to the differential equation dz/dt = 2te^z that passes through the origin is z = ln(1/(-t^2 - 1)).