Question

1) yy'-e^x= 0, and y=4 when x=0 means that:
A)y=x-lnx^2+4
B)y^2=4x^2+3
C)y=-2x+1/2x^3
D)y^2=2e^x+14
E)1/2ln|x^2+4|+6

Answer 1

If you rewrite y' and treat 'dy' and 'dx' as individual parts, you can separate this equation out with algebra into one part having only ys and the other having only xs. \[y \frac{dy}{dx}-e^x=0\] So it should look like an integral for y on the left and an integral for x on the right. Try it out.

Answer 2

D)y^2=2e^x... ?

Answer 3

You are exactly correct but do not forget to write the integration constant "C" in the RHS of the equation.

Then with the given values of x and y, you can find out C.

Answer 4

? so you still need to do more work after D?

Answer 5

What if there were no options ?

What if I gave you these options:

(a) y^2 = 2e^x + 4
(b) y^2 = 2e^x
(c) y^2 = 2e^x + 14
(d) etc

Answer 6

I thought it was 2e^x+14, but IDK

Answer 7

Its all right.

You must understand that evaluating the integration constant is a very important step in solving differential equations.

Your equation at present is like this:

y^2 = 2e^x + C

Answer 8

Now, it is given that when x = 0, y = 4.

So, put x = 0 and y = 4 in (y^2 = 2e^x + C) and find out the value of C.

Answer 9

4^2=2e^0+c 16=2(1)+c c=14?

Answer 10

Very good

Answer 11

Now, your answer would be complete :

y^2 = 2e^x + 14

Answer 12

ok, thank you. I get it now.

Answer 13

Good work

Answer 14

@vishweshshrimali5 - Can you explain how to do this one? I didn't get it.

Answer 15

We have : \(y \cfrac{dy}{dx} = e^x\)

Are we required to integrate both sides? or something else?

@vishweshshrimali5

Answer 16

Yep, basically you're just multiplying "dx" on both sides and the dx divides out on the left side leaving an integral in y. Both integrals will have an arbitrary constant, but you can combine them into one on either side.