Question

d/dx ln(ax+b) .solve the following​

Answer 1

Explanation:

First Order

They are "First Order" when there is only dydx , not d2ydx2 or d3ydx3 etc

Linear

A first order differential equation is linear when it can be made to look like this:

dydx + P(x)y = Q(x)

Where P(x) and Q(x) are functions of x.

To solve it there is a special method:

We invent two new functions of x, call them u and v, and say that y=uv.

We then solve to find u, and then find v, and tidy up and we are done!

And we also use the derivative of y=uv (see Derivative Rules (Product Rule) ):

dydx = udvdx + vdudx

Steps

Here is a step-by-step method for solving them:

1. Substitute y = uv, and

dydx = udvdx + vdudx

into

dydx + P(x)y = Q(x)

2. Factor the parts involving v

3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)

4. Solve using separation of variables to find u

5. Substitute u back into the equation we got at step 2

6. Solve that to find v

7. Finally, substitute u and v into y = uv to get our solution!

Let's try an example to see:

Example 1: Solve this:

  dydx − yx = 1

First, is this linear? Yes, as it is in the form

dydx + P(x)y = Q(x)

where P(x) = −1x and Q(x) = 1

So let's follow the steps:

Step 1: Substitute y = uv, and   dydx = u dvdx + v dudx

So this:dydx − yx = 1

Becomes this:udvdx + vdudx − uvx = 1

Step 2: Factor the parts involving v

Factor v:u dvdx + v( dudx − ux ) = 1

Step 3: Put the v term equal to zero

v term equal to zero:dudx − ux = 0

So:dudx = ux

Step 4: Solve using separation of variables to find u

Separate variables:duu = dxx

Put integral sign:∫ duu = ∫ dxx

Integrate:ln(u) = ln(x) + C

Make C = ln(k):ln(u) = ln(x) + ln(k)

And so:u = kx

Answer 2

hope it helps u...........