How to Solve the Differential Equation y'' + y = tan^2(x)?

[Mark Brewer]

Homework Statement

y'' + y = tan²x

The Attempt at a Solution

y_h = c₁sinx + c₂cosx

y_p = -y₁ ∫ (y₂r/W) dx + y₂ ∫ (y₁r/W) dx

r = tan²x

y₁ = sinx

y₂ = cosx

W = 1

I'm using integration by parts, but I've realized that the trig functions repeat (trivial).

Is there another way to solve this equation or is by parts the way to go?

 

[ehild]

Show your integrals, please. What have you tried?

 

[Mark Brewer]

Show your integrals, please. What have you tried?

Thank you for the reply.

I retried by parts and saw that trig identities simplify the integrals, so I don't have to do by parts a second time. I apologize for not displaying my integrals.

My answer is y = c₁sinx + c₂cos +cosx

if you still want to see my work I can display it, but it's a lot to type. Let me know.

 

[ehild]

Thank you for the reply.

I retried by parts and saw that trig identities simplify the integrals, so I don't have to do by parts a second time. I apologize for not displaying my integrals.

My answer is y = c₁sinx + c₂cos +cosx

if you still want to see my work I can display it, but it's a lot to type. Let me know.

Is that really solution of the differential equation?
I would like to see your integrals and what you have tried.

 

[Mark Brewer]

W = 1
r = tan²x
y₁ = sinx
y₂ = cosx

y_h = c₁sinx + c₂cosx

y_p = -y₁ ∫ ((y₂)(r)/W)dx + y₂ ∫ ((y₁)(r)/W)dx

y_p = -sinx ∫ ((cosx)(tan²x)dx) + cosx ∫ ((sinx)(tan²x)dx)

y_p = -sinx ∫ ((cosx)(sec²x +1)dx) + cosx ∫ ((sinx)(sec²x +1)dx)

There's two by parts labeled 1 (left side) and 2 (right side)

by parts 1

u = cosx , du = sinx dx
v = sec²x +1 , dv = tanx + x

-sinx(cosx)(tanx +x) - ∫ (sinxtanx - sinx)dx
-sinx(cosx)(tanx +x) + ∫ ((sin²x/cosx) - sinx)dx
-sinx(cosx)(tanx +x) + ∫ (((cos²x + 1(/(cosx) + sinx)dx
-sinx(cosx)(tanx +x) + ∫ (((cos²x + 1(/(cosx) - sinx)dx
-sinx(cosx)(tanx +x) + ∫ (1/cosx)dx + ∫(cosx)dx - ∫(sinx)dx
-sinxcosxtanx + (sinxcosx)x + ln cosx + sinx - cosx
-sin²x + (sinxcosx)x +ln cosx + sinx - cosx

by parts 2

u = cosx , du = sinx dx
v = sec²x +1 , dv = tanx + x

+ cosx(sinx)(tanx + x) -∫ (cosx)(tanx + x)dx
+ cosx(sinx)(tanx + x) -∫ (sinx)dx + ∫(cosx)dx
+ cosx(sinx)(tanx + x) +cosx + sinx
+ cosxsinxtanx + (cosxsinx)x + cosx + sinx
+ sin²x + (cosxsinx)x + cosx + sinx

y = c₁sinx + c₂cosx -sin²x + (sinxcosx)x +ln cosx + sinx - cosx+ sin²x + (cosxsinx)x + cosx + sinx

Final answer:

y = c₁sinx + c₂cosx + (sin²xcos²x)x² +ln cosx + 2sinx

Thanks for having me type out this problem I found mistakes. How does this answer look?

 

[ehild]

W = 1

y_p = -sinx ∫ ((cosx)(tan²x)dx) + cosx ∫ ((sinx)(tan²x)dx)

y_p = -sinx ∫ ((cosx)(sec²x +1)dx) + cosx ∫ ((sinx)(sec²x +1)dx)

tan²x is not the same as sec²x +1. Check. But I would not use the secant. Would do it with tan(x)=sin(x)/cos(x) instead.
I do not understand your way of integration by parts, either.
And W=-1.

 

[Mark Brewer]

tan²x is not the same as sec²x +1. Check. But I would not use the secant. Would do it with tan(x)=sin(x)/cos(x) instead.
I do not understand your way of integration by parts, either.

Thank you, the secant should have been sec²x -1

 

[ehild]

Thank you, the secant tan² should have been sec²x -1

 

[Mark Brewer]

Hi Ehild,

My apologies for thinking I could use sec²x - 1. I thought the derivation was tanx from sec²x. I'll use sinx/cosx.

Thank you again for the help.

 

[Mark Brewer]

My new answer is extremely long.

y = c₁sinx + c₂cosx + sin²x - xcosx(xsinx - 1) - lntanx + secx - cosx(sinxtanx - 2) -xsinx(cosx + 1)

 

[ehild]

My new answer is extremely long.

y = c₁sinx + c₂cosx + sin²x - xcosx(xsinx - 1) - lntanx + secx - cosx(sinxtanx - 2) -xsinx(cosx + 1)

Still not correct, I am afraid. Check the parentheses. I can not tell where your mistakes are if you do not show your work.
You need the integrals $\int(\cos(x)\tan^2(x)dx)$ and $\int(\sin(x)\tan^2(x)dx)$ . How did you do them?Hint: when integrating by parts, integrate u'=sin(x).