- #1
ollyfinn
- 14
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Hi I have not studied calculus for a while and I am just seeking some clarification on the following two problems I have attempted to solve.
PROBLEM 1
dy/dx = y(x^3 - √x)
I have separated the variables as follows:
Rewrote equation as dy/dx = y(x^3 - x^1/2)
Divided both sides by 1/y
1/y dy/dx = x^3 - x^1/2
Multiplied both sides by dx
1/y dy = x^3 - x^1/2 dx
Then to find the indefinite integral
∫1/y dy = ∫x^3 - x^1/2 dx
ln[y] = 0.25x^4 - 1.5x^3/2
y = e^(0.25x^4 - 1.5x^3/2) + C
Does this look correct?
PROBLEM 2
(yx^4 - yx^2) = dy/dx y^2.x^4
Separating variables:
Multiply both sides by dx
(yx^4 - yx^2) dx = y^2.x^4 dy
Divide both sides by y
x^4 - x^2 dx = y.x^4 dy
Divide boths sides by x^4
(x^4 - x^2)/x^4 dx = y dy
Swap LHS and RHS so
y dy = (x^4 - x^2)/x^4 dx
Then to find the indefinite integral
∫y dy = ∫(x^4 - x^2)/x^4 dx
y^2/2 = (x^2 + 1)/x + C
Solve for y
y = √2(x^2 + 1)/x + C
If my solutions are incorrect then any pointers in the right direction would be gratefully received.
Thanks
PROBLEM 1
dy/dx = y(x^3 - √x)
I have separated the variables as follows:
Rewrote equation as dy/dx = y(x^3 - x^1/2)
Divided both sides by 1/y
1/y dy/dx = x^3 - x^1/2
Multiplied both sides by dx
1/y dy = x^3 - x^1/2 dx
Then to find the indefinite integral
∫1/y dy = ∫x^3 - x^1/2 dx
ln[y] = 0.25x^4 - 1.5x^3/2
y = e^(0.25x^4 - 1.5x^3/2) + C
Does this look correct?
PROBLEM 2
(yx^4 - yx^2) = dy/dx y^2.x^4
Separating variables:
Multiply both sides by dx
(yx^4 - yx^2) dx = y^2.x^4 dy
Divide both sides by y
x^4 - x^2 dx = y.x^4 dy
Divide boths sides by x^4
(x^4 - x^2)/x^4 dx = y dy
Swap LHS and RHS so
y dy = (x^4 - x^2)/x^4 dx
Then to find the indefinite integral
∫y dy = ∫(x^4 - x^2)/x^4 dx
y^2/2 = (x^2 + 1)/x + C
Solve for y
y = √2(x^2 + 1)/x + C
If my solutions are incorrect then any pointers in the right direction would be gratefully received.
Thanks