Final answer to the problem
$e^{\frac{3}{200}t^2}x=\frac{3}{4}\sum_{n=0}^{\infty } \frac{\left(\frac{3}{200}\right)^nt^{\left(2n+1\right)}}{\left(2n+1\right)\left(n!\right)}+C_0$
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Step-by-step Solution
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- Exact Differential Equation
- Linear Differential Equation
- Separable Differential Equation
- Homogeneous Differential Equation
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Integrate by substitution
- Integrate by parts
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1
Rearrange the differential equation
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$\frac{dx}{dt}-\frac{-3xt}{100}=\frac{3}{4}$
Learn how to solve problems step by step online. Solve the differential equation dx/dt=3/4+(-3xt)/100. Rearrange the differential equation. Simplifying. We can identify that the differential equation has the form: \frac{dy}{dx} + P(x)\cdot y(x) = Q(x), so we can classify it as a linear first order differential equation, where P(t)=\frac{3t}{100} and Q(t)=\frac{3}{4}. In order to solve the differential equation, the first step is to find the integrating factor \mu(x). To find \mu(t), we first need to calculate \int P(t)dt.
Final answer to the problem
$e^{\frac{3}{200}t^2}x=\frac{3}{4}\sum_{n=0}^{\infty } \frac{\left(\frac{3}{200}\right)^nt^{\left(2n+1\right)}}{\left(2n+1\right)\left(n!\right)}+C_0$
