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Can this differential equation be solved using separation of variables? (dy)/(dx)=(cos(x))/(3y-y^(2)) Choose 1 answer: (A) Yes (B) No

Can this differential equation be solved using separation of variables?\newlinedydx=cos⁡(x)3y−y2 \frac{d y}{d x}=\frac{\cos (x)}{3 y-y^{2}} \newlineChoose 11 answer:\newline(A) Yes\newline(B) No

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Intermediate Value Theoremright chevron icon

Full solution

Q. Can this differential equation be solved using separation of variables?\newlinedydx=cos⁡(x)3y−y2 \frac{d y}{d x}=\frac{\cos (x)}{3 y-y^{2}} \newlineChoose 11 answer:\newline(A) Yes\newline(B) No
  1. Check Equation: First, we need to check if the differential equation can be rearranged so that all terms involving xx are on one side and all terms involving yy are on the other side. This is the essence of separation of variables.
  2. Multiply by (3y−y2)(3y - y^2): We start by multiplying both sides of the equation by (3y−y2)(3y - y^2) to move the y terms to the left side:\newline(3y−y2)dydx=cos⁡(x)(3y - y^2)\frac{dy}{dx} = \cos(x).
  3. Divide by (3y−y2)(3y - y^2): Next, we divide both sides by (3y−y2)(3y - y^2) to isolate dydy on the left side:\newlinedy=(cos⁡(x)3y−y2)dx.dy = \left(\frac{\cos(x)}{3y - y^2}\right) dx.
  4. Isolate dydy: Now, we have successfully separated the variables, with all yy terms on the left side and all xx terms on the right side. This means that the differential equation can be solved using the method of separation of variables.

More problems from Intermediate Value Theorem

Question
Let f f be a continuous function on the closed interval [1,5] [1,5] , where f(1)=1 f(1)=1 and f(5)=−3 f(5)=-3 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=2 f(c)=2 for at least one c c between −3-3 and 11\newline(B) f(c)=−2 f(c)=-2 for at least one c c between −3-3 and 11\newline(C) f(c)=2 f(c)=2 for at least one c c between 11 and 55\newline(D) f(c)=−2 f(c)=-2 for at least one c c between 11 and 55
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Posted 9 months ago

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Let h h be a continuous function on the closed interval [−3,4] [-3,4] , where h(−3)=−1 h(-3)=-1 and h(4)=2 h(4)=2 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) h(c)=1 h(c)=1 for at least one c c between −3-3 and 44\newline(B) h(c)=−2 h(c)=-2 for at least one c c between −3-3 and 44\newline(C) h(c)=−2 h(c)=-2 for at least one c c between −1-1 and 22\newline(D) h(c)=1 h(c)=1 for at least one c c between −1-1 and 22
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Let g g be a continuous function on the closed interval [−3,3] [-3,3] , where g(−3)=0 g(-3)=0 and g(3)=6 g(3)=6 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) g(c)=−2 g(c)=-2 for at least one c c between −3-3 and 33\newline(B) g(c)=−2 g(c)=-2 for at least one c c between 00 and 66\newline(C) g(c)=5 g(c)=5 for at least one c c between −3-3 and 33\newline(D) g(c)=5 g(c)=5 for at least one c c between 00 and 66
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Let g g be a continuous function on the closed interval [−1,4] [-1,4] , where g(−1)=−4 g(-1)=-4 and g(4)=1 g(4)=1 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) g(c)=3 g(c)=3 for at least one c c between −1-1 and 44\newline(B) g(c)=−3 g(c)=-3 for at least one c c between −4-4 and 11\newline(C) g(c)=3 g(c)=3 for at least one c c between −4-4 and 11\newline(D) g(c)=−3 g(c)=-3 for at least one c c between −1-1 and 44
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Question
Let g g be a continuous function on the closed interval [−1,3] [-1,3] , where g(−1)=−2 g(-1)=-2 and g(3)=−5 g(3)=-5 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) g(c)=−3 g(c)=-3 for at least one c c between −5-5 and −2-2\newline(B) g(c)=0 g(c)=0 for at least one c c between −5-5 and −2-2\newline(C) g(c)=0 g(c)=0 for at least one c c between −1-1 and 33\newline(D) g(c)=−3 g(c)=-3 for at least one c c between −1-1 and 33
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Let f f be a continuous function on the closed interval [−2,1] [-2,1] , where f(−2)=3 f(-2)=3 and f(1)=6 f(1)=6 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=0 f(c)=0 for at least one c c between −2-2 and 11\newline(B) f(c)=0 f(c)=0 for at least one c c between 33 and 66\newline(C) f(c)=4 f(c)=4 for at least one c c between −2-2 and 11\newline(D) f(c)=4 f(c)=4 for at least one c c between 33 and 66
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Let f f be a continuous function on the closed interval [−5,0] [-5,0] , where f(−5)=0 f(-5)=0 and f(0)=5 f(0)=5 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=−2 f(c)=-2 for at least one c c between −5-5 and 00\newline(B) f(c)=2 f(c)=2 for at least one c c between 00 and 55\newline(C) f(c)=2 f(c)=2 for at least one c c between −5-5 and 00\newline(D) f(c)=−2 f(c)=-2 for at least one c c between 00 and 55
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Posted 9 months ago

Question
Let f f be a continuous function on the closed interval [1,5] [1,5] , where f(1)=1 f(1)=1 and f(5)=−3 f(5)=-3 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=2 f(c)=2 for at least one c c between 11 and 55\newline(B) f(c)=−2 f(c)=-2 for at least one c c between −3-3 and 11\newline(C) f(c)=−2 f(c)=-2 for at least one c c between 11 and 55\newline(D) f(c)=2 f(c)=2 for at least one c c between −3-3 and 11
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Posted 9 months ago

Question
Let h(x)=1x h(x)=\frac{1}{x} .\newlineCan we use the intermediate value theorem to say the equation h(x)=0.5 h(x)=0.5 has a solution where −1≤x≤1 -1 \leq x \leq 1 ?\newlineChoose 11 answer:\newline(A) No, since the function is not continuous on that interval.\newline(B) No, since 00.55 is not between h(−1) h(-1) and h(1) h(1) .\newline(C) Yes, both conditions for using the intermediate value theorem have been met.
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Posted 9 months ago

Question
Let h h be a continuous function on the closed interval [0,4] [0,4] , where h(0)=2 h(0)=2 and h(4)=−2 h(4)=-2 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) h(c)=3 h(c)=3 for at least one c c between 00 and 44\newline(B) h(c)=−1 h(c)=-1 for at least one c c between 00 and 44\newline(C) h(c)=−1 h(c)=-1 for at least one c c between −2-2 and 22\newline(D) h(c)=3 h(c)=3 for at least one c c between −2-2 and 22
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