2 Questions Involving Calculus (Differential Equations)?

Question

Greetings and thank you in advance for reading my question.

At the moment I've started to get a grasp of how to tackle differential equations (that being separate variables, integrate, solve for C, substitute and rearrange), but these two questions have stumped me for the past 2 hours

The two questions are 
1) Find the solution of the differential equation that satisfies the given initial condition. 
y'(x) = sqrt[-2(y(x))+17] , y(5) = 4 
y(x) = ? 
& 
2) What is the equation of the curve whose y-intercept is 4 and it's slope at any point is (dy/dx) = 9yx^2 
y(x) = ?

My Attempts for: 
1) y(x) = sqrt[(x^2)/2-42)] , y(x) = sqrt[(x^2)/2 -8] 
2) y(x) = 4.5x +6 
(all answers were wrong but I thought I'd at-least show that I tried to solve them)

Any help would be greatly appreciated, whether that be a hint, a guideline or a step in the right direction.

? · Accepted Answer

Note that in your 2 questions, the DE's are separable. So separate them and integrate!
Do this here and we will take it from there.

For 2), if y=y(x) is your sought curve, they are telling you that its slope  (dy/dx) is equal to 9yx^2 .

Some Body · Answer

1)dy/dx = √(-2y + 17)

Separating:
dy / √(-2y + 17) = dx
(-2y + 17)^(-1/2) dy = dx

u substitution:
u = -2y + 17
du = -2dy
dy = -1/2 du

u^(-1/2) (-1/2 du) = dx
-1/2 u^(-1/2) du = dx

Integrate:
-1/2 (2u^(1/2)) = x + C
-√u = x + C
√u = -x + C

Substitute back:
√(-2y + 17) = -x + C

Initial condition y(5) = 4:
√(-2*4 + 17) = -5 + C
3 = -5 + C
C = 8

Therefore:
√(-2y + 17) = -x + 8
You can solve for y if you wish.

2) dy/dx = 9yx^2

Separating:
dy / y = 9x^2 dx

Integrating:
ln(y) = 3x^3 + C

Initial condition y(0) = 4:
ln(4) = 0 + C
C = ln(4)

Therefore:
ln(y) = 3x^3 + ln(4)

Again, you can solve for y if you wish.

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2 Questions Involving Calculus (Differential Equations)?

2 Answers