Q. Solve the equation.dxdy=4x3y−6x2yChoose 1 answer:(A) y=ex4−2x3+C(B) y=±x4−2x3+C(C) y=Cex4−2x3(D) y=Cx4−2x3
Identify Equation Type: This is a first-order linear homogeneous differential equation. We can solve it by separating variables or by recognizing it as a separable differential equation. We will use the separation of variables method.
Separate Variables: First, we separate the variables y and x by dividing both sides of the equation by y and by 4x3−6x2.ydy=4x3−6x24x3−6x2dx
Simplify Right Side: We simplify the right side of the equation by factoring out 2x2 from both terms.ydy=2x2(2x−3)2x2(2x−3)dx
Integrate Both Sides: After canceling out the common terms, we get: (ydy)=dx
Exponentiate to Solve: Now we integrate both sides of the equation. The integral of y1dy is ln∣y∣, and the integral of dx is x. ∫(y1)dy=∫dxln∣y∣=x+C
Check Answer Choices: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm.eln∣y∣=ex+C∣y∣=ex⋅eC
Check Answer Choices: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm.e(ln∣y∣)=e(x+C)∣y∣=ex⋅eC Since eC is just a constant, we can denote it as C′, and since y can be positive or negative, we can drop the absolute value and include the constant C′ to account for the sign.y=C′ex
Check Answer Choices: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm.e(ln∣y∣)=e(x+C)∣y∣=ex⋅eCSince eC is just a constant, we can denote it as C′, and since y can be positive or negative, we can drop the absolute value and include the constant C′ to account for the sign.y=C′exWe now look at the answer choices to see which one matches our solution.(A)y=e(x4−2x3)+C is not correct because it does not match the form of our solution.(B)y=±x4−2x3+C is not correct because it involves a square root and our solution does not.(C)y=Ce(x4−2x3) is not correct because the exponent does not match our solution.(D)y=Cx4−2x3 is not correct because it involves a square root and multiplication by x3, which does not match our solution.
Check Answer Choices: To solve for y, we exponentiate both sides of the equation to get rid of the natural logarithm.e(ln∣y∣)=e(x+C)∣y∣=ex⋅eCSince eC is just a constant, we can denote it as C′, and since y can be positive or negative, we can drop the absolute value and include the constant C′ to account for the sign.y=C′exWe now look at the answer choices to see which one matches our solution.(A)y=e(x4−2x3)+C is not correct because it does not match the form of our solution.(B)y=±x4−2x3+C is not correct because it involves a square root and our solution does not.(C)y=Ce(x4−2x3) is not correct because the exponent does not match our solution.(D)y=Cx4−2x3 is not correct because it involves a square root and multiplication by x3, which does not match our solution.None of the provided answer choices match our solution y=C′ex. It seems there might be a mistake in the separation of variables step or in the provided answer choices.
More problems from Find derivatives of using multiple formulae