Final answer to the problem
$e^{2t^2}x=-5\sum_{n=0}^{\infty } \frac{2^nt^{\left(2n+1\right)}}{\left(2n+1\right)\left(n!\right)}+C_0$
Got another answer? Verify it here!
Step-by-step Solution
How should I solve this problem?
- Choose an option
- 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
- Load more...
Can't find a method? Tell us so we can add it.
1
Group the terms of the equation
Learn how to solve problems step by step online.
$\frac{dx}{dt}+4xt=-5$
Learn how to solve problems step by step online. Solve the differential equation dx/dt+4xt+5=0. Group the terms of the equation. 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)=4t and Q(t)=-5. 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. So the integrating factor \mu(t) is.
Final answer to the problem
$e^{2t^2}x=-5\sum_{n=0}^{\infty } \frac{2^nt^{\left(2n+1\right)}}{\left(2n+1\right)\left(n!\right)}+C_0$
