Differential equation problem?

Question

Solve the differential equation.

e^x dy/dx + 4e^x y = 3, x > 0

cos x dy/dx + y sin x = sin x cos x

hey listen, im trying to figure out these problems for a good half an hour now. if u dunno how to do it, then u dont need to comment lol.

Answer 1

you shouldn't be in a differentials class if you can't do the problems....do your own homework or drop out

Answer 2

e^x dy/dx + 4e^x y = 3, x > 0

Use a substitution:
e^x y = u and du/dx = e^x y + e^x dy/dx
du/dx = u + e^x dy/dx
dy/dx = (du/dx - u)/e^x

(du/dx - u) + 4u = 3
du/dx = 3(1 - u)
du/(1 - u) = 3dx
-ln(1 - u) = 3x + c ..... c is an arbitrary constant
u = 1 - e^(-3x - c) = e^x y
y = [1 - e^(-3x - c)]/e^x

check:
dy/dx = [-(1 - e^(-3x - c)) + 3e^(-3x - c))] /e^x

substitute this and y into the original equation:
[-(1 - e^(-3x - c)) + 3 e^(-3x - c))] + 4[ [1 - e^(-3x - c)]/] = 3
and 3 = 3 so the solution checks out


cos x dy/dx + y sin x = sin x cos x

u = ysecx and du/dx = secx dy/dx + ysecxtanx
dy/dx = cosx du/dx - ytanx
cosx(cosx du/dx - ytanx) + ysinx = sinx cosx
cos^2x du/dx - ysinx + ysinx = sinxcosx
du/dx = tanx
u = ln(secx) + c ..... c is an arbitrary constant
ysecx = ln(secx) + c
y = cosx ln(secx) + c*cosx

Check:
dy/dx = - sinx ln(secx) + cosx tan x- c*sinx

substitute this and y into the original equation:
cosx(-sinx ln(secx) +sin x- c*sinx) + sinx(cosx ln(secx) + c*cosx) = sinx cosx
And this is right

Answer 3

for this one: cos x dy/dx + y sin x = sin x cos x

you multiply 1/cosx to every term, therefore getting this eqn:

dy/dx + y(sinx/cosx) = sinx ----> eqn 1

sinx/cosx is the same as tanx so you replace tan x to the said eqn, getting:

dy/dx + ytanx = sinx ----> eqn 2

As observed, this equation is linear in y, which is in the form

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

where P(x) = tanx and Q(x) = sinx.

Get the integrating factor "phi" by this eqn:

Φ = e^[∫P(x) dx]
Φ = e^[∫tanx dx]
= e^(ln|secx|)
Φ = secx

so the general solution for this eqn is given by:

yΦ = ∫ΦQ(x) dx

Substituting Φ= secx,

ysecx = ∫secx (sinx) dx + C
ysecx = ∫ tanxdx + C

ysecx = ln|secx| + C.

hope this helps. :) :) :)

Answer 4

This is a linear D.E. Arrange in this form. y' + 8y = 32 p(x) = 8; q(x) = 32 I.F. = e^integral of p(x)dx I.F. = e^8x Therefore: ye^8x = integral(32e^8xdx) + C ye^8x = 32/8 e^8x + C ye^8x = 4e^8x + C y = 4 + C e^(-8x); then y = 7 as x = 0 7 = 4 + Ce^0 C = 3 So, y = 4 +3e^(-8x)...end

Answer 5

dy/dx + ytanx = sinx
dy/dx=sinx-ytanx
dy=1/y(sinx/y-tanx)dx
now interage this
and consider x/y some this as

Answer 6

1♠ thus y’+4y = 3exp(-x);
this is Bernoulli: y’ +p(x)*y =q(x);
y= exp(-P(x)) *[∫q(x)* exp(P(x)) *dx +C],
where P(x)=∫p(x)*dx =4x; thus
♣ y= exp(-4x) *[3∫exp(-x)* exp(4x) *dx +C]=
= exp(-4x) *[exp(3x) +C]=exp(-x) +C*exp(-4x);

2♠ thus y’ +tan(x)*y = sin x;
y= exp(-P(x)) *[∫q(x)* exp(P(x)) *dx +C]
where P(x)=∫tan(x)*dx =-ln|cos x|; thus
♣ y= exp(ln|cos x|) *[∫(sinx)* exp(-ln|cos x|) *dx +C]=
= cosx *[∫( sinx/cos x) *dx +C]=
= -cosx *ln|cos x| +C*cos x;