Tricky Maths Q. Please Help!?

Question

Can anybody please solve these differential equations. Im stumped. Even hints would be appreciated. Cheers

1) dy/dx = -2y when y(0) = 6

2) dy/dx = (X^2 + 1) e^-y when y(1) = 0

Answer 1

1)first,rearrange the equation
1/ydy= -2 dx

integrate both sides
lny = -2x+C.......(1)
where C is a constant
we now want to find the value of C
from the initial conditions given
y(0)=6,that is,when x=0,y=6
plug these values for x and y
into (1)
ln6= -2*0+C
>>C=ln6=1.791759469
substitute back into (1)
lny= -2x+ln6
raise e to the power of each side
e^lny=e^(-2x+ln6)
>>y=e^(ln6)*e^(-2x)
>>>y=6*e^(-2x)

{we will now substitute back into
original equation to see if we
are correct
dy/dx= -12e^(-2x) from answer
-2y= -12e^(-2x) from answer
hence,answer is correct}

2)first,rearrange the equation
e^ydy=x^2+1

integrate both sides
e^y=x^3/3+x+C..........(2)
y(1)=0,that is y=0 when x=1
substitute into (2)
1=1/3+1+C
>>>>C= -1/3
substitute back into (2)
e^y=x^3/3+x-1/3.......(3)
take ln of each side

y=ln{(x^3+3x-1)/3}

{we now substitute back into
the original equation to see if
we are correct
from answer,
dy/dx= (x^2+1)*1/((x^3+3x-1)/3)
=(x^2+1)*e^(-y) from (3)
hence,the answer is correct}

i hope that you understand my
working

Answer 2

4

Answer 3

1) dy/dx = -2y
Integral of (1/y) dy = Integral of -2 dx
lny = -2x + C

When x=0 and y=6, ln6 = -2(0) + C => C = ln6

So lny = -2x + ln6

2) dy/dx = (X^2 + 1) e^-y
Integral of e^y dy = Integral of (x^2 + 1) dx
e^y = (1/3)x^3 + x + C

When x=1 and y=0,
e^0 = (1/3)(1^3) + 1 + C
1 = 1/3 + 1 + C
C = -1/3

So e^y = (1/3)x^3 + x - 1/3

Answer 4

He He

Answer 5

im not doing your hw for you.

Answer 6

ouch! i dont envy you!