Ordinary Differential Equations
If possible please show me in detail

7.15

This is a Bernoulli equation in disguise. Solve for :

Divide through both sides by :

Substitute and , and the resulting ODE is linear in .

Multiply both sides by , integrate both sides, solve for , then for :

8.15

This is separable; write , then

Integrate both sides to get

Given that when , we find

so the particular solution is

10.15

The image is a bit blurry, but it looks like the equation is

Substitute , so that and the equation is linear in :

Divide through both sides by :

Integrate both sides and solve for :

Solve for by integrating both sides twice:

11.15

Unfortunately, I'm not sure how to tackle this one just yet...

12.15

You could use undetermined coefficients here, but since the equation is free of , you can substitute and to reduce the order and get a linear equation.

Multiply through both sides by , integrate, solve ... you know the drill:

13.15

We could reduce the order again, but that won't avoid having to solve a higher-order ODE like in the previous examples. The homogeneous equation

has characteristic equation

with roots and (with multiplicity 2), and thus the characteristic solution is

For the particular solution, assume an ansatz

with derivatives

Substitute these into the ODE and simplify to end up with

Matching up coefficients tells us

So the particular solution is

and the general solution to the ODE is

14.15

The characteristic equation is

with roots , which admits the characteristic solution

Assume the ansatz

with second derivative

Substitute these into the ODE, simplify, and solve for :

Then the general solution is

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)