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CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q25: Solve dydx=x3cosec⁡y\frac{dy}{dx} = x^3\operatorname{cosec} y⁠, given y(0)=0y(0) = 0

Sample 2020-21Ch 9 · Differential Equations
The question

Solve the following differential equation:

dydx=x3cosec⁡y\dfrac{dy}{dx} = x^3 \operatorname{cosec} y⁠, given that y(0)=0y(0) = 0⁠.

Exam answer
dydx=x3cosec⁡y\frac{dy}{dx} = x^3\operatorname{cosec} y

Separating the variables,

dycosec⁡y=x3 dx\frac{dy}{\operatorname{cosec} y} = x^3\,dx
Why?

The right side is a product of a function of xx and a function of yy⁠, so the variables can be separated. Divide both sides by cosec⁡y\operatorname{cosec} y and multiply by dxdx⁠: everything with yy goes with dydy⁠, everything with xx goes with dxdx⁠.

Integrating both sides,

∫dycosec⁡y=∫x3 dx⇒∫sin⁡y dy=∫x3 dx⇒−cos⁡y=x44+C\begin{aligned} \int\frac{dy}{\operatorname{cosec} y} &= \int x^3\,dx \\ \Rightarrow \int\sin y\,dy &= \int x^3\,dx \\ \Rightarrow -\cos y &= \frac{x^4}{4} + C \end{aligned}
Why?

Since cosec⁡y=1sin⁡y\operatorname{cosec} y = \frac{1}{\sin y}⁠, dividing by it is the same as multiplying by sin⁡y\sin y⁠.

Standard integrals:

∫sin⁡y dy=−cos⁡y∫x3 dx=x3+13+1=x44\begin{aligned} \int\sin y\,dy &= -\cos y \\ \int x^3\,dx &= \frac{x^{3+1}}{3+1} = \frac{x^4}{4} \end{aligned}

One constant CC on one side is enough. Watch out for the sign: the derivative of cos⁡y\cos y is −sin⁡y-\sin y⁠, so the integral of sin⁡y\sin y is −cos⁡y-\cos y⁠.

Given y=0y = 0 when x=0x = 0⁠: −cos⁡0=0+C-\cos 0 = 0 + C⁠, so C=−1C = -1⁠.

Why?

The condition y(0)=0y(0) = 0 fixes the constant and turns the general solution into the particular one. Since cos⁡0=1\cos 0 = 1⁠, the left side is −1-1 and the right side is 044+C=C\frac{0^4}{4} + C = C⁠.

−cos⁡0=044+C−1=0+CC=−1\begin{aligned} -\cos 0 &= \frac{0^4}{4} + C \\ -1 &= 0 + C \\ C &= -1 \end{aligned}
−cos⁡y=x44−1-\cos y = \frac{x^4}{4} - 1
Why?

Put C=−1C = -1 into −cos⁡y=x44+C-\cos y = \frac{x^4}{4} + C⁠. Then multiply both sides by −1-1 to get cos⁡y\cos y on its own.

−cos⁡y=x44−1cos⁡y=−x44+1(×−1)=1−x44\begin{aligned} -\cos y &= \frac{x^4}{4} - 1 \\ \cos y &= -\frac{x^4}{4} + 1 \quad (\times -1) \\ &= 1 - \frac{x^4}{4} \end{aligned}

∴ cos⁡y=1−x44\therefore\ \cos y = 1 - \dfrac{x^4}{4} is the required solution.

Where this comes from

Chapter 9, Differential Equations: Variables separable: particular solution.

Question type: Variables separable + A particular solution (combines 2 types)

How to spot it: click to see

Type 3. Variables separable: yy on one side, xx on the other

Spot it: dydx\frac{dy}{dx} equals a part in xx times a part in yy⁠, like (1+x2)(1+y2)(1 + x^2)(1 + y^2)⁠.

Here: x3cosec⁡yx^3\operatorname{cosec}y is a part in xx times a part in yy⁠. Move the cosec⁡y\operatorname{cosec}y across, as sin⁡y dy=x3 dx\sin y\,dy = x^3\,dx⁠, and integrate both sides.

Type 4. A particular solution: solve, then find CC

Spot it: a condition such as "y=1y = 1 when x=0x = 0", or "the curve passes through a point".

Here: Then y(0)=0y(0) = 0 gives −cos⁡0=0+C-\cos 0 = 0 + C⁠, so C=−1C = -1 and cos⁡y=1−x44\cos y = 1 - \frac{x^4}{4}⁠.

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