Differential Equations — Flashcards
Telangana Open School (TOSS) · Class 12 · Mathematics
25 flashcards for Differential Equations (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms and facts.
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Find the order and degree of the differential equation: $$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 = x$$
Answer
Step 1: Identify the highest derivative → $$\frac{d^2y}{dx^2}$$ → Order = 2. Step 2: The power of the highest derivative is 1 (no exponent written means power 1). So, Degree = 1. Answer: Order = 2,…
What is the order and degree of: $$\left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 = 0$$
Answer
Step 1: The highest derivative is $$\frac{d^3y}{dx^3}$$ → Order = 3. Step 2: This derivative is squared → power is 2 → Degree = 2. Answer: Order = 3, Degree = 2.
Find the order and degree of: $$\frac{dy}{dx} = \sin\left(\frac{d^2y}{dx^2}\right)$$
Answer
Step 1: The highest derivative is $$\frac{d^2y}{dx^2}$$ → Order = 2. Step 2: The sine function contains the second derivative, but it's not raised to a power. The equation is not polynomial in deriva…
Determine the order and degree of: $$\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = k\frac{d^2y}{dx^2}$$
Answer
Step 1: Highest derivative is $$\frac{d^2y}{dx^2}$$ → Order = 2. Step 2: The equation has a fractional power (3/2). To find degree, remove radicals by squaring both sides: $$\left[1 + \left(\frac{dy…
Is the differential equation $$\frac{dy}{dx} + y^2 = x$$ linear or non-linear? Why?
Answer
The equation is **non-linear**. Reason: The dependent variable 'y' is raised to the power 2 in the term $$y^2$$. In a linear differential equation, the dependent variable and its derivatives must app…
Why is $$\frac{d^2y}{dx^2} + y\frac{dy}{dx} = 0$$ non-linear?
Answer
This equation is **non-linear** because the term $$y\frac{dy}{dx}$$ involves the product of the dependent variable 'y' and its derivative $$\frac{dy}{dx}$$. In a linear differential equation, no such…
Solve: $$\frac{dy}{dx} = 3x^2$$
Answer
This is a simple separable equation. Step 1: Write as: $$dy = 3x^2 dx$$ Step 2: Integrate both sides: $$\int dy = \int 3x^2 dx$$ $$y = 3 \cdot \frac{x^3}{3} + C$$ $$y = x^3 + C$$ Answer: $$y = x…
Solve: $$\frac{dy}{dx} = e^{x} \cdot y$$
Answer
Step 1: Separate variables: $$\frac{dy}{y} = e^x dx$$ Step 2: Integrate both sides: $$\int \frac{1}{y} dy = \int e^x dx$$ $$\ln|y| = e^x + C$$ Step 3: Solve for y: $$|y| = e^{e^x + C} = e^C \cdo…
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