Question

Find the solution of the differential equation that satisfies the given initial conditions?

(Just give me tips, you don't have to do the whole thing)

a. `yy'=x\sinx` and `y(0)=-1`
b. `y'\tanx=y+8`, `y(\pi/3)=8` and `0\ltx\lt\pi/2`
c. `xy'=y+x^2\sinx` and `y(\pi)=0`

Answer 1

`(a) \ y^2 = 2sinx -2xcosx + 1`
`(b) \ y = (32sqrt(3))/3 sinx - 8`
`(c) \ y = -xcosx -x`

Explanation:

Part (a):

`yy'=xsinx` with `y(0)=-1`

This is separable, so by "separating the variables" , we get:

`int \ y \ dy = int \ xsinx \ dx` ..... [A]

To integrate the RHS we apply IBP

Let # { (u,=x, => (du)/dx,=1), ((dv)/dx,=sinx, => v,=-cosx ) :}#

Then plugging into the IBP formula:

`int \ u (dv)/dx \ dx = uv - int \ v (dv)/dx \ dx`

So we have:

`int \ xsinx \ dx = -xcosx + int \ cosx \ dx = sinx -xcosx`

Thus if we integrate [A] we get:

`1/2 y^2 = sinx -xcosx + C`

Applying the initial conditions:

`y(0)=-1=> 1/2 = 0 -0 + C`

Thus we gain the particular solution

`1/2 y^2 = sinx -xcosx + 1/2`

`:. y^2 = 2sinx -2xcosx + 1`

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Part (b):

`y' tanx = y+8` with `y(pi/3)=8` and `0 lt x lt pi/2`

This is also separable:

`1/(y+8)y' = cotx`

We can "separate the variables" to get:

`int \ 1/(y+8) \ dy = int \ cotx \ dx`

Which is trivial to integrate:

`ln|y+8| = ln|sinx| + C`

Applying the initial conditions:

`y(pi/3)=8 => ln|8+8| = ln|sin (pi/3)| + C`
`:. ln16 = ln(sqrt(3)/2) + C => C = ln((32sqrt(3))/3)`

Thus we gain the particular solution:

`ln|y+8| = ln|sinx| + ln((32sqrt(3))/3)`
`:. y+8 = (32sqrt(3))/3 sinx`
`:. y = (32sqrt(3))/3 sinx - 8`

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Part (c):

`xy' = y + x^2 sinx` with `y(pi)=0`

We can write as:

`y' - y/x = x sinx`

Which is of the form:

`dy/dx + P(x)y=Q(x)`

So, we can construct an Integrating Factor:

`I = exp(int \ P(x) \ dx`
`\ \ = exp(int \-1/x \ dx)`
`\ \ = exp(-ln(x))`
`\ \ = 1/x`

And if we multiply the DE by this Integrating Factor, `I`, we will have a perfect product differential;

`1/x \ y' - 1/x^2 \ y = 1/x \ x sinx`

`:. d/dx (1/x \ y) = sinx`

This is now separable, so we can "separate the variables" to get:

`y/x = int \ sinx \ dx`

Which is trivial to integrate:

`y/x = -cosx + C`

Applying the initial conditions:

`y(pi)=0 => 0 = -cospi + C => C = -1`

Thus we gain the particular solution:

`y/x = -cosx -1`

`:. y = -xcosx -x`