Solve these differential equations

Solve the linear equation x ( dy(x))/( dx) - (x + 1) y(x) = -e^x (x^2 + 1):
Rewrite the equation:
( dy(x))/( dx) + ((-x - 1) y(x))/x = -(e^x (x^2 + 1))/x
Let μ(x) = e^( integral(-x - 1)/x dx) = e^(-x)/x

Multiply both sides by μ(x):
(e^(-x) ( dy(x))/( dx))/x + ((e^(-x) (-x - 1)) y(x))/x^2 = -(x^2 + 1)/x^2
Substitute (e^(-x) (-x - 1))/x^2 = ( d)/( dx)(e^(-x)/x):
(e^(-x) ( dy(x))/( dx))/x + ( d)/( dx)(e^(-x)/x) y(x) = -(x^2 + 1)/x^2
Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:

( d)/( dx)((e^(-x) y(x))/x) = -(x^2 + 1)/x^2
Integrate both sides with respect to x:
 integral( d)/( dx)((e^(-x) y(x))/x) dx = integral-(x^2 + 1)/x^2 dx

Evaluate the integrals:
(e^(-x) y(x))/x = -x + 1/x + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = e^(-x)/x:
Answer: |y(x) = e^x x (-x + 1/x + C1)

Solve the linear equation (y(x))/sqrt(x^2 + 1) + ( dy(x))/( dx) = 1:
Let μ(x) = e^( integral1/sqrt(x^2 + 1) dx) = e^(sinh^(-1)(x)).

Multiply both sides by μ(x):
e^(sinh^(-1)(x)) ( dy(x))/( dx) + (e^(sinh^(-1)(x)) y(x))/sqrt(x^2 + 1) = e^(sinh^(-1)(x))

Substitute e^(sinh^(-1)(x))/sqrt(x^2 + 1) = ( d)/( dx)(e^(sinh^(-1)(x))):
e^(sinh^(-1)(x)) ( dy(x))/( dx) + ( d)/( dx)(e^(sinh^(-1)(x))) y(x) = e^(sinh^(-1)(x))

Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(e^(sinh^(-1)(x)) y(x)) = e^(sinh^(-1)(x))

Integrate both sides with respect to x:
 integral( d)/( dx)(e^(sinh^(-1)(x)) y(x)) dx = integral e^(sinh^(-1)(x)) dx

Evaluate the integrals:
e^(sinh^(-1)(x)) y(x) = 1/2 (sinh^(-1)(x) + x (x + sqrt(x^2 + 1))) + c_1, where c_1 is an arbitrary constant.

Divide both sides by μ(x) = e^(sinh^(-1)(x)):
Answer: |y(x) = (1/2 (sinh^(-1)(x) + x (x + sqrt(x^2 + 1))) + C1)/e^(sinh^(-1)(x))

Get the answers by entering them into Wolfram/Alpha Computing Engine. If you need step by step solution, I will borrow my brother's computer and might be able to help you, since he is subscribed to it.

Hi Majid,

I am not going to attempt your calculus but it is great to finally see you posting on the forum  :D

I always like looking at these answers too if anyone want to have a go. :)

y'+y* tanx=1/cosx

y' + y [tan x]  = sec x          multiply both sides by sec x

y'sec x  + y [  sec x tan x ]   =  sec^2 x

Let u  =  y sec x   →  du   =  y'sec x  + y*  sec x tan x

So

du  = sec^2 x     integrate both sides

∫ du   = ∫ sec^2 dx

u =  tan x + C₁

y sec x   =  tan x + C₁   

y [ 1/cos x ]   = sin x / cos x  + C₁      multiply through by cos x

y =   sin x + C₁ cos x

xy'+2y=x²-3        multiply both sides by x

x^2y'+2xy=x^3-3x

Let u  =  x^2*y   →  du   =  x^2y'  + 2xy

So we have

du  = x^3-3x      integrate both sides

∫ du   = ∫ x^3 -3x dx

u  = x^4/4  - (3/2)x^2  + C₁

x^2*y = x^4/4  - (3/2)x^2  + C₁     divide through by   x^2

y  =  x^2/4  - (3/2)  +  C₁ / x^2

y'+6y/(x+2)=1/(x+2)²   multiply both sides by (x+ 2)^6

y' (x + 2)^6  +  6y(x + 2)^5  =  (x + 2)^4

Let u   = y (x + 2)^6    →  du =   y' (x + 2)^6  +  6y(x + 2)^5  

So we have

du   = (x + 2)^4    integrate both sides

∫ du   = ∫  (x + 2)^4 dx

u =  (x + 2)^5 / 5    +  C₁

y (x + 2)^6  =  (x + 2)^5 / 5    +  C₁        divide through by   ( x + 2)^6

y  =   1 / [5( x+ 2)]  +  C₁ / ( x + 2)^6

y  =   1 / [5x + 10]  +  C₁ / ( x + 2)^6

I think that it's intended that they are all solved using the so called Integrating Factor (IF) Method.

The standard form of the equation is taken to be

\(\displaystyle \frac{dy}{dx}+P(x)y = Q(x)\) ,

and the method of solution is to multiply throughout by the IF  \(\displaystyle e^{\int P(x)\,dx}\). That will put the LHS into the form of a perfect derivative.

For example, with the second one, \(P(x)=\tan(x) \text{ so the IF will be } \exp(\ln(\sec(x))=\sec(x)\).

That's what CPhill multiplied by. (I don't know whether he worked it out as above or whether it was by inspection).

For the first one, \(\displaystyle x\frac{dy}{dx} +2y=x^{2}-3\),

divide throughout by x to begin with to put the equation into its standard form, so we have

\(\displaystyle \frac{dy}{dx}+\frac{2}{x}y = x -\frac{3}{x}\) and the IF will be \(\displaystyle \exp(\int \frac{2}{x}\;dx)=\exp(2\ln(x))=\exp(\ln(x^{2}))=x^{2}\).

Multiply throughout by \(\displaystyle x^{2}\) and the equation becomes

\(\displaystyle x^{2}\frac{dy}{dx}+2xy=x^{3}-3x\)

which can be written as

\(\displaystyle \frac{d}{dx}(x^{2}y)=x^{3}-3x\),

and now just integrate wrt x.

The third one also begins with a division by x.

The fourth one requires a knowledge of hyperbolic functions. The IF will be \(\displaystyle e^{\;\sinh^{-1}x}\) and we have to integrate that on the RHS.

To do that, use the substitution \(\displaystyle u = \sinh^{-1}x\) so that \(\displaystyle x = \sinh(u)\) and then later write \(\displaystyle \cosh(u)=(e^{u}+e^{-u})/2\).

It's still messy after that (putting it back in terms of x).

Tiggsy

Thanks, Tiggsy......!!!!!