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Solve the equation.

(dy)/(dx)=4x^(3)y-6x^(2)y
Choose 1 answer:
(A) 
y=e^(x^(4)-2x^(3))+C
(B) 
y=+-sqrt(x^(4)-2x^(3)+C)
(C) 
y=Ce^(x^(4)-2x^(3))
(D) 
y=Csqrt(x^(4)-2)x^(3)

Solve the equation.\newlinedydx=4x3y6x2y \frac{d y}{d x}=4 x^{3} y-6 x^{2} y \newlineChoose 11 answer:\newline(A) y=ex42x3+C y=e^{x^{4}-2 x^{3}}+C \newline(B) y=±x42x3+C y= \pm \sqrt{x^{4}-2 x^{3}+C} \newline(C) y=Cex42x3 y=C e^{x^{4}-2 x^{3}} \newline(D) y=Cx42x3 y=C \sqrt{x^{4}-2} x^{3}

Full solution

Q. Solve the equation.\newlinedydx=4x3y6x2y \frac{d y}{d x}=4 x^{3} y-6 x^{2} y \newlineChoose 11 answer:\newline(A) y=ex42x3+C y=e^{x^{4}-2 x^{3}}+C \newline(B) y=±x42x3+C y= \pm \sqrt{x^{4}-2 x^{3}+C} \newline(C) y=Cex42x3 y=C e^{x^{4}-2 x^{3}} \newline(D) y=Cx42x3 y=C \sqrt{x^{4}-2} x^{3}
  1. 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.
  2. Separate Variables: First, we separate the variables yy and xx by dividing both sides of the equation by yy and by 4x36x24x^3 - 6x^2.dyy=4x36x24x36x2dx\frac{dy}{y} = \frac{4x^3 - 6x^2}{4x^3 - 6x^2}dx
  3. Simplify Right Side: We simplify the right side of the equation by factoring out 2x22x^2 from both terms.\newlinedyy=2x2(2x3)dx2x2(2x3)\frac{dy}{y} = \frac{2x^2(2x - 3)dx}{2x^2(2x - 3)}
  4. Integrate Both Sides: After canceling out the common terms, we get: (dyy)=dx(\frac{dy}{y}) = dx
  5. Exponentiate to Solve: Now we integrate both sides of the equation. The integral of 1ydy\frac{1}{y} \, dy is lny\ln|y|, and the integral of dxdx is xx. \newline(1y)dy=dx\int \left(\frac{1}{y}\right) dy = \int dx\newlinelny=x+C\ln|y| = x + C
  6. Check Answer Choices: To solve for yy, we exponentiate both sides of the equation to get rid of the natural logarithm.elny=ex+Ce^{\ln|y|} = e^{x + C}y=exeC|y| = e^x \cdot e^C
  7. Check Answer Choices: To solve for yy, we exponentiate both sides of the equation to get rid of the natural logarithm.\newlinee(lny)=e(x+C)e^{(\ln|y|)} = e^{(x + C)}\newliney=exeC|y| = e^x \cdot e^C Since eCe^C is just a constant, we can denote it as CC', and since yy can be positive or negative, we can drop the absolute value and include the constant CC' to account for the sign.\newliney=Cexy = C'e^x
  8. Check Answer Choices: To solve for yy, we exponentiate both sides of the equation to get rid of the natural logarithm.e(lny)=e(x+C)e^{(\ln|y|)} = e^{(x + C)}y=exeC|y| = e^x \cdot e^CSince eCe^C is just a constant, we can denote it as CC', and since yy can be positive or negative, we can drop the absolute value and include the constant CC' to account for the sign.y=Cexy = C'e^xWe now look at the answer choices to see which one matches our solution.(A) y=e(x42x3)+C(A)\ y=e^{(x^{4}-2x^{3})}+C is not correct because it does not match the form of our solution.(B) y=±x42x3+C(B)\ y=\pm\sqrt{x^{4}-2x^{3}+C} is not correct because it involves a square root and our solution does not.(C) y=Ce(x42x3)(C)\ y=Ce^{(x^{4}-2x^{3})} is not correct because the exponent does not match our solution.(D) y=Cx42x3(D)\ y=C\sqrt{x^{4}-2}x^{3} is not correct because it involves a square root and multiplication by x3x^3, which does not match our solution.
  9. Check Answer Choices: To solve for yy, we exponentiate both sides of the equation to get rid of the natural logarithm.e(lny)=e(x+C)e^{(\ln|y|)} = e^{(x + C)}y=exeC|y| = e^x \cdot e^CSince eCe^C is just a constant, we can denote it as CC', and since yy can be positive or negative, we can drop the absolute value and include the constant CC' to account for the sign.y=Cexy = C'e^xWe now look at the answer choices to see which one matches our solution.(A) y=e(x42x3)+C(A)\ y=e^{(x^{4}-2x^{3})}+C is not correct because it does not match the form of our solution.(B) y=±x42x3+C(B)\ y=\pm\sqrt{x^{4}-2x^{3}+C} is not correct because it involves a square root and our solution does not.(C) y=Ce(x42x3)(C)\ y=Ce^{(x^{4}-2x^{3})} is not correct because the exponent does not match our solution.(D) y=Cx42x3(D)\ y=C\sqrt{x^{4}-2}x^{3} is not correct because it involves a square root and multiplication by x3x^3, which does not match our solution.None of the provided answer choices match our solution y=Cexy = C'e^x. It seems there might be a mistake in the separation of variables step or in the provided answer choices.

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