Algebra
Calculus
Trigonometry
Matrix
Question
Solve the differential equation
y=3e(x3)C+e(x3),C∈R
Evaluate
dxdy+3x2y=x2
Move the expression to the right side
dxdy=x2−3x2y
Rewrite the expression
dxdy=x2(1−3y)
Rewrite the expression
1−3y1×dxdy=x2
Transform the expression
1−3y1×dy=x2dx
Integrate the left-hand side of the equation with respect to y and the right-hand side of the equation with respect to x
∫1−3y1dy=∫x2dx
Calculate
More Steps
Evaluate
∫1−3y1dy
Use the property of integral ∫ax+b1dx=a1ln(ax+b)
−31ln(−3y+1)
Use b−a=−ba=−ba to rewrite the fraction
−31ln(−3y+1)
Add the constant of integral C1
−31ln(−3y+1)+C1,C1∈R
−31ln(−3y+1)+C1=∫x2dx,C1∈R
Calculate
More Steps
Evaluate
∫x2dx
Use the property of integral ∫xndx=n+1xn+1
2+1x2+1
Add the numbers
2+1x3
Add the numbers
3x3
Add the constant of integral C2
3x3+C2,C2∈R
−31ln(−3y+1)+C1=3x3+C2,C1∈R,C2∈R
Since the integral constants C1 and C2 are arbitrary constants, replace them with constant C
−31ln(−3y+1)=3x3+C,C∈R
Calculate
More Steps
Evaluate
−31ln(−3y+1)=3x3+C
Change the sign
31ln(−3y+1)=−3x3+C
Multiply by the reciprocal
31ln(−3y+1)×3=(−3x3+C)×3
Multiply
ln(−3y+1)=(−3x3+C)×3
Multiply
More Steps
Evaluate
(−3x3+C)×3
Apply the distributive property
−3x3×3+C×3
Reduce the numbers
−x3+C×3
Since C is a constant,replace the C×3 with the constant C
−x3+C
ln(−3y+1)=−x3+C
Convert the logarithm into exponential form using the fact that logax=b is equal to x=ab
−3y+1=e−x3+C
Move the constant to the right-hand side and change its sign
−3y=e−x3+C−1
Change the signs on both sides of the equation
3y=−e−x3+C+1
Divide both sides
33y=3−e−x3+C+1
Divide the numbers
y=3−e−x3+C+1
y=3−e−x3+C+1,C∈R
Rewrite the expression
More Steps
Evaluate
−e−x3+C
Use am+n=am×an to expand the expression
−eC×e−x3
Since the expression eC is a constant,it is possible to denote that whole expression as a constant C
−Ce−x3
Since C is a constant,replace the C(−1) with the constant C
Ce−x3
y=3Ce−x3+1,C∈R
Solution
y=3e(x3)C+e(x3),C∈R
Show Solution
