Answer :
Sure, let's solve the differential equation step by step. The equation is:
[tex]\[ \frac{(2t^2 + 3t)}{(2s + s)} \frac{ds}{dt} = \frac{t}{s+1} \][/tex]
First, let's simplify the equation. Note that [tex]\(2s + s = 3s\)[/tex], so the equation becomes:
[tex]\[ \frac{2t^2 + 3t}{3s} \frac{ds}{dt} = \frac{t}{s+1} \][/tex]
Now, we can separate variables. Multiply both sides by [tex]\(3s(s+1)\)[/tex] to get:
[tex]\[ (2t^2 + 3t) \, ds = 3s \cdot t \, dt \][/tex]
Now we can integrate both sides separately. Rewrite the equation in a separated form:
[tex]\[ \int \frac{3s}{2t^2 + 3t} \, ds = \int t(s+1) \, dt \][/tex]
We need to solve these integrals:
### Left Integral:
Let's integrate the left side with respect to [tex]\(s\)[/tex]:
[tex]\[ \int \frac{3s}{2t^2 + 3t} \, ds \][/tex]
### Right Integral:
For the right side, we integrate with respect to [tex]\(t\)[/tex]:
[tex]\[ \int t(s+1) \, dt \][/tex]
Let's focus on integrating these properly.
### Simplify by separable variables:
Separate and integrate as follows:
[tex]\[ \int \frac{1}{s} \, ds = \int \frac{t}{2t^2 + 3t} \, dt + \int \frac{1}{2t^2 + 3t} \, dt \][/tex]
For the simplicity purpose, we need to look at the integrals:
[tex]\[ \int \frac{1}{s} \, ds = \ln|s| + C_1 \][/tex]
Similarly, rewrite the integral on the right-hand side by separating as follows:
[tex]\[ \int \left( \frac{t}{2t^2 + 3t} + \frac{1}{2t^2 + 3t} \right) \, dt \][/tex]
We notice that the first integral can be solved by direct observation and use of logarithm properties:
### Integrate with partial fraction decomposition:
[tex]\[ \int t \left( \frac{1}{t(2t + 3)} \right) \, dt + \int \left( \frac{1}{2t(t + \frac{3}{2})} \right) \, dt \][/tex]
Hence the right-hand integral becomes:
### Substitution on left integral:
Using u-substitution where [tex]\(u = 2t^2+3t \)[/tex]:
[tex]\[ \int \frac t{u} du = \frac{\ln|u|}{2} \][/tex]
Integrate right-hand...
So, combining all:
[tex]\(\ln|s| = \frac{\ln|2t^2+ 3t|}{2} \)[/tex]
Solving the equation, we get
[tex]\(\boxed{s(t)} = e^{\frac{\ln|u|}{2}}}= \sqrt{u}} Boundary conditionsir \(|u | = |2t^2+3t|\)[/tex]
Hence \(\sqrt(u) }= sqrt(2t^3+ 3t
With constant `C`
[tex]\[ \frac{(2t^2 + 3t)}{(2s + s)} \frac{ds}{dt} = \frac{t}{s+1} \][/tex]
First, let's simplify the equation. Note that [tex]\(2s + s = 3s\)[/tex], so the equation becomes:
[tex]\[ \frac{2t^2 + 3t}{3s} \frac{ds}{dt} = \frac{t}{s+1} \][/tex]
Now, we can separate variables. Multiply both sides by [tex]\(3s(s+1)\)[/tex] to get:
[tex]\[ (2t^2 + 3t) \, ds = 3s \cdot t \, dt \][/tex]
Now we can integrate both sides separately. Rewrite the equation in a separated form:
[tex]\[ \int \frac{3s}{2t^2 + 3t} \, ds = \int t(s+1) \, dt \][/tex]
We need to solve these integrals:
### Left Integral:
Let's integrate the left side with respect to [tex]\(s\)[/tex]:
[tex]\[ \int \frac{3s}{2t^2 + 3t} \, ds \][/tex]
### Right Integral:
For the right side, we integrate with respect to [tex]\(t\)[/tex]:
[tex]\[ \int t(s+1) \, dt \][/tex]
Let's focus on integrating these properly.
### Simplify by separable variables:
Separate and integrate as follows:
[tex]\[ \int \frac{1}{s} \, ds = \int \frac{t}{2t^2 + 3t} \, dt + \int \frac{1}{2t^2 + 3t} \, dt \][/tex]
For the simplicity purpose, we need to look at the integrals:
[tex]\[ \int \frac{1}{s} \, ds = \ln|s| + C_1 \][/tex]
Similarly, rewrite the integral on the right-hand side by separating as follows:
[tex]\[ \int \left( \frac{t}{2t^2 + 3t} + \frac{1}{2t^2 + 3t} \right) \, dt \][/tex]
We notice that the first integral can be solved by direct observation and use of logarithm properties:
### Integrate with partial fraction decomposition:
[tex]\[ \int t \left( \frac{1}{t(2t + 3)} \right) \, dt + \int \left( \frac{1}{2t(t + \frac{3}{2})} \right) \, dt \][/tex]
Hence the right-hand integral becomes:
### Substitution on left integral:
Using u-substitution where [tex]\(u = 2t^2+3t \)[/tex]:
[tex]\[ \int \frac t{u} du = \frac{\ln|u|}{2} \][/tex]
Integrate right-hand...
So, combining all:
[tex]\(\ln|s| = \frac{\ln|2t^2+ 3t|}{2} \)[/tex]
Solving the equation, we get
[tex]\(\boxed{s(t)} = e^{\frac{\ln|u|}{2}}}= \sqrt{u}} Boundary conditionsir \(|u | = |2t^2+3t|\)[/tex]
Hence \(\sqrt(u) }= sqrt(2t^3+ 3t
With constant `C`
