Solve the initial value problem. (dy)/(dx) + 6y

Transcript

00:01 In this problem we are as to find the solution of the initial value problem and it is nothing but d .y by d x minus y is equal to 2x e to the power 2x with the initial condition y of 0 is equal to 3.

00:21 Now first of all let's compare the given differential equation with an equation day y by d x plus plus p y is equal to q.

00:33 If we compare these two equations, we can find the value of p as minus 1 and the value of q as 2x times e to the power 2x.

00:46 Now we know that the integrating factor for such cases is equal to e to the power integral of p d x.

00:56 In our case, the integrating factor will be equal to e to the power integral log.

01:03 The value of p is minus 1.

01:05 So minus 1 d x.

01:07 If we integrate minus 1 with respect to x, we will have minus x.

01:13 So we will have e to the power minus x as our integrating factor.

01:21 Now recall that the required solution of the given differential equal.

01:27 Equation is nothing but y times the integrating factor is equal to integral of q times integrating factor times dx plus c.

01:43 Now let's substitute all the known values in this expression.

01:49 So we will have y times e to the power minus x the integrating factor is equal to integral of q is 2x, e to the power 2x times integrating factor is e to the power minus x d x plus c and it will be equal to integral of 2x e to the power x d x plus c...

Question

Solve the initial value problem.\\ $\frac{dy}{dx} + 6y - 3e^{-3x} = 0$, $y(0) = -2$\ The solution is $y(x) = 3e^{-3x} - e$.

Question

Solve the initial value problem.\\ $\frac{dy}{dx} + 6y - 3e^{-3x} = 0$, $y(0) = -2$\ The solution is $y(x) = 3e^{-3x} - e$.

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