(x^2+2x+10)dy/dx=2x^3-x^2-4x-1, how do I solve this ODE?

Question

I haven't seen this kind of separable ODE before, where do I start.

Answer 1

The equation can be written: (x^2+2x+1+9)dy/dx=(x+1)(2x^2-3x-1). This is: ((x+1)^2+9)dy/dx=(x+1)(2x^2-3x-1). Let x+1=3tan(a), so dx/da=3sec^2(a) and dy/da=dy/dx*dx/da. Replacing x+1 with 3tan(a), we have (9tan^2(a)+9)dy/dx=3tan(a)(2(3tan(a)-1)^2-3(3tan(a)-1)-1). So, since 3sec^2(a)dy/dx=dy/da, and 1+tan^2(a)=sec^2(a), we can write:

3dy/da=3tan(a)(18tan^2(a)-12tan(a)+2-9tan(a)+3-1). So dy/da=tan(a)(18tan^2(a)-21tan(a)+4). Expanding the right side we get dy/da=18tan^3(a)-21tan^2(a)+4tan(a). The expression on the right can be integrated, and it becomes more obvious when sec^2(a)=1+tan^2(a) is substituted. tan^3(a)=tan(a)(sec^2(a)-1), so integral(tan^3(a)da)=integral((tan(a)sec^2(a)-tan(a))da). Also tan^2(a)=sec^2(a)-1, so -21tan^2(a)=21-21sec^2(a), the integral wrt a of which is 21a-21tan(a).

Let z=tan(a), then dz=sec^2(a)da. So the first part of the integral becomes integral(zdz)=z^2/2=tan^2(a)/2. Now let z=cos(a), so dz=-sin(a)da. Integral(-tan(a)da) becomes integral(-da.sin(a)/cos(a))=integral(dz/z)=ln(z)=ln(cos(a)). The whole integral now becomes integral(18tan^3(a)-21tan^2(a)+4tan(a))

=9tan^2(a)+18ln(cos(a))+21a-21tan(a)-4ln(cos(a))=9tan^2(a)+14ln(cos(a))+21a-21tan(a). We now replace tan(a)=(x+1)/3 or a=tan^-1((x+1)/3). Here goes:

y=(x+1)^2+14ln(sqrt(x^2+2x+10)/3)+21tan^-1((x+1)/3)-7(x+1)+constant

or

y=(x+1)^2+7ln((x^2+2x+10)/9)+21tan^-1((x+1)/3)-7(x+1)+constant

 

Answer 2

Given:(x²+2x+10)dy=(2x³-x²-4x+1)dx   x²+2x+10=(x+1/2)²+(39/4)>0, so we have:

dy/dx=(2x³-x²-4x+1/(x²+2x+10) ··· Eq.1

Let y=∫(dy/dx)dx＝∫[(2x³-x²-4x+1)/(x²+2x+10)]dx ··· Eq.2   While,

2x³-x²-4x+1=(2x-5)(x²+2x+10)-7(2x+2)*+65, so Eq.1 can be rewritten:

dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10), so Eq.2 also can be rewritten:

y=∫(2x-5)dx-7∫[(2x+2)/(x²+2x+10)]dx+65∫[1/(x²+2x+10)]dx ··· Eq.3   use integration formula for Eq.3

(i). ∫[f'(x)/f(x)]dx=lnlf(x)l, so ∫[(2x+2)/(x²+2x+10)]dx=ln(x²+2x+10)+C1

(ii). In a quadratic equation ax²+bx+c=0, if the discriminant D=b²-4ac is negative(<0),

∫[1/(ax²+bx)]dx=(2/√-D)arctan[(2ax+b)/√-D]+C2.   Here, D=2²-4·1·10=-36<0, so

∫[1/(x²+2x+10)]dx=(2/√36)arctan[(2x+2)/√-36]=(1/3)arctan[(x+1)/3] +C3   Thus, Eq.3 can be rewritten:

y=x²-5x-7ln(x²+2x+10)+(65/3)arctan[(x+1)/3]+C ··· Eq.4

CK: Differentiate each term of Eq.4 using derivative rules such as (x^n)'=nx^(n-1), (ln x)'=1/x, (arctanx)'=1/(1+x²) and chainrules.

(x²-5x)'=2x-5, -7[ln(x²+2x+10)]'=-7[(2x+2)/(x²+2x+10)], and 65/3[arctan(x+1)/3]'=65/(x²+2x+10) So, the first derivative Eq.4 is:

y'=dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10)=[(2x+5)(x²+2x+10)-7(2x+2)+65}/(x²+2x+10)

=(2x³-x²-4x+1)/(x²+2x+10) (=Eq.1)   CKD.

The answer is: y=∫(dy/dx)dx=x²-5x-7ln(x²+2x+10)+(65/3)arctan(x+1)/3+C

Answer 3

Given:(x²+2x+10)dy/dx=2x³-x²-4x+1   x²+2x+10=(x+1/2)²+(39/4)>0, so we have:

dy/dx=(2x³-x²-4x+1)/(x²+2x+10) ··· Eq.1

Let y=∫(dy/dx)dx=∫[(2x³-x²-4x+1)/(x²+2x+10)]dx ··· Eq.2   While,

2x³-x²-4x+1=(2x-5)(x²+2x+10)-7(2x+2)*+65, so Eq.1 can be rewritten:

dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10), so Eq.2 also can be rewritten:

y=∫(2x-5)dx-7∫[(2x+2)/(x²+2x+10)]dx+65∫[1/(x²+2x+10)]dx ··· Eq.3   Use formulas for Eq.3.

(i). ∫[f'(x)/f(x)]dx=lnf(x), so ∫[(2x+2)*/(x²+2x+10)]dx=ln(x²+2x+10)+C1

(ii). In a quadratic equation ax²+bx+c=0, if the discriminant D=b²-4a2xc is negative(<0), then

∫[1/(ax²+bx+c)]dx=(2/√-D)arctan[(2ax+b)/√-D]+C1   Here, D=2²-4·1·10=-36<0, so √-D=√36=6, and

∫[1/(x²+2x+10)dx=(1/3)arctan[(x+1)/3]+C3   Thus, Eq.3 can be rewritten:

y=x²-5x-7ln(x²+2x+10)+(65/3)arctan[(x+1)/3]+C ··· Eq.4

CK: Differentiate each term of Eq.4: (x²-5x)'=2x-5, -7[ln(x²+2x+10)]'=-7[(2x+2)/(x²+2x+10)], and

(65/3)[arctan(x+1)/3]'=65/(x²+2x+10)   So, the first derivative of Eq.4 is:

y'=dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10)=(2x³-x²-4x+1)/(x²2x+10)=Eq.1  CKD.

The answer: y=x²-5x-7ln(x²+2x+10)+(65/3)arctan[(x+1)/3]+C