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Separable differential equations

Math • Differential Equations • Knowledge-Based • 5 questions

Separable differential equations Knowledge-Based Download PDF

Math · Calculus · Math · Differential Equations

5 Math practice questions on Separable differential equations. Questions that assess core concepts and understanding.

Questions

  1. 1.

    What is the particular solution to the differential equation dydx=2xyx2+1\frac{dy}{dx} = \frac{2xy}{x^2 + 1} satisfying the initial condition y(1)=6y(1) = 6?

    • A) y=x2+5y = x^2 + 5
    • B) y=3x2+3y = 3x^2 + 3
    • C) y=6ex2−1y = 6e^{x^2 - 1}
    • D) y=12x2+1y = \frac{12}{x^2 + 1}
  2. 2.

    What is the particular solution to the differential equation dydx=2x+3y2\frac{dy}{dx} = \frac{2x + 3}{y^2} satisfying the initial condition y(0)=2y(0) = 2?

    • A) y=3x2+9x+83y = \sqrt[3]{3x^2 + 9x + 8}
    • B) y=3x2+9x+23y = \sqrt[3]{3x^2 + 9x + 2}
    • C) y=x2+3x+4y = \sqrt{x^2 + 3x + 4}
    • D) y=x2+3x+83y = \sqrt[3]{x^2 + 3x + 8}
  3. 3.

    What is the particular solution to the differential equation dydx=cos⁡xey\frac{dy}{dx} = \frac{\cos x}{e^y} satisfying the initial condition y(0)=0y(0) = 0?

    • A) y=ln⁡(cos⁡x)y = \ln(\cos x)
    • B) y=sin⁡xy = \sin x
    • C) y=ln⁡(sin⁡x)y = \ln(\sin x)
    • D) y=ln⁡(sin⁡x+1)y = \ln(\sin x + 1)
  4. 4.

    What is the particular solution to the differential equation dydx=2x+12y\frac{dy}{dx} = \frac{2x + 1}{2y} satisfying the initial condition y(1)=2y(1) = 2 for y>0y > 0?

    • A) y=x2+x+4y = \sqrt{x^2 + x + 4}
    • B) y=x2+x+2y = \sqrt{x^2 + x + 2}
    • C) y=2x2+x+1y = \sqrt{2x^2 + x + 1}
    • D) y=x2+x+2y = x^2 + x + 2
  5. 5.

    What is the particular solution to the differential equation dydx=2x(y2+1)\frac{dy}{dx} = 2x(y^2 + 1) satisfying the initial condition y(0)=1y(0) = 1?

    • A) y=tan⁡(x2+1)y = \tan(x^2 + 1)
    • B) y=tan⁡(2x2+π4)y = \tan\left(2x^2 + \frac{\pi}{4}\right)
    • C) y=2e2x2−1y = \sqrt{2e^{2x^2} - 1}
    • D) y=tan⁡(x2+π4)y = \tan\left(x^2 + \frac{\pi}{4}\right)