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Calculus Examples
1ydx+3ydy=01ydx+3ydy=0
Step 1
Subtract 1ydx1ydx from both sides of the equation.
3ydy=-1ydx3ydy=−1ydx
Step 2
Multiply both sides by yy.
y(3y)dy=y(-1y)dxy(3y)dy=y(−1y)dx
Step 3
Step 3.1
Rewrite using the commutative property of multiplication.
3y⋅ydy=y(-1y)dx3y⋅ydy=y(−1y)dx
Step 3.2
Multiply yy by yy by adding the exponents.
Step 3.2.1
Move yy.
3(y⋅y)dy=y(-1y)dx3(y⋅y)dy=y(−1y)dx
Step 3.2.2
Multiply yy by yy.
3y2dy=y(-1y)dx3y2dy=y(−1y)dx
3y2dy=y(-1y)dx3y2dy=y(−1y)dx
Step 3.3
Rewrite using the commutative property of multiplication.
3y2dy=-y1ydx3y2dy=−y1ydx
Step 3.4
Cancel the common factor of yy.
Step 3.4.1
Factor yy out of -y−y.
3y2dy=y⋅-11ydx3y2dy=y⋅−11ydx
Step 3.4.2
Cancel the common factor.
3y2dy=y⋅-11ydx
Step 3.4.3
Rewrite the expression.
3y2dy=-1dx
3y2dy=-1dx
3y2dy=-1dx
Step 4
Step 4.1
Set up an integral on each side.
∫3y2dy=∫-1dx
Step 4.2
Integrate the left side.
Step 4.2.1
Since 3 is constant with respect to y, move 3 out of the integral.
3∫y2dy=∫-1dx
Step 4.2.2
By the Power Rule, the integral of y2 with respect to y is 13y3.
3(13y3+C1)=∫-1dx
Step 4.2.3
Simplify the answer.
Step 4.2.3.1
Rewrite 3(13y3+C1) as 3(13)y3+C1.
3(13)y3+C1=∫-1dx
Step 4.2.3.2
Simplify.
Step 4.2.3.2.1
Combine 3 and 13.
33y3+C1=∫-1dx
Step 4.2.3.2.2
Cancel the common factor of 3.
Step 4.2.3.2.2.1
Cancel the common factor.
33y3+C1=∫-1dx
Step 4.2.3.2.2.2
Rewrite the expression.
1y3+C1=∫-1dx
1y3+C1=∫-1dx
Step 4.2.3.2.3
Multiply y3 by 1.
y3+C1=∫-1dx
y3+C1=∫-1dx
y3+C1=∫-1dx
y3+C1=∫-1dx
Step 4.3
Apply the constant rule.
y3+C1=-x+C2
Step 4.4
Group the constant of integration on the right side as K.
y3=-x+K
y3=-x+K
Step 5
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
y=3√-x+K