Solve the following differential equation: dy dx =y -2 - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$

✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}=-2\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\tan\text{x}$
$\text{Q}=-2\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\tan\text{xdx}}$
$=\text{e}^{-\log|\sec\text{x}|}=\cos\text{x}$
Multiplying both sides of (1) by $\cos\text{x},$ we get
$\cos\text{x}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\tan\text{x}\Big)=-2\sin\text{x}\times\cos\text{x}$
$\Rightarrow\ \cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=-\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\cos\text{x}=-\int\sin2\text{x dx + C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{\cos2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=\frac{1-2\sin^2\text{x}}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\frac{1}{2}+\text{C}$
$\Rightarrow\ \text{y}\cos\text{x}=-\sin^2\text{x}+\text{K}$ $\Big($where $\text{K}=\frac{1}2+\text{c}\Big)$
$\Rightarrow\ \text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$
Hence, $\text{y}=\sec\text{x}\big(-\sin^2\text{x}+\text{K}\big)$ is the required solution.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(5 Marks)" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $AD$ is the median of $\triangle\text{ABC},$ using vectors, prove that $\text{AB}^2+\text{AC}^2=2\big(\text{AD}^2+\text{CD}^2\big).$

Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$

Show that the points whose position vectors are $-2\hat{\text{i}}+3\hat{\text{j}},\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $7\hat{\text{i}}-\hat{\text{k}}$ are collinear.

Find the equation of the normal to the curve $x^2 + 2y^2 - 4x - 6y + 8 = 0$ at the point whose abscissa is $2.$

Find the equation of the plane passing through the intersection of the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})=7,\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$ and the point (2, 1, 3).

There are two types of fertilizers $F_1$ and $F_2 . F_1$ consists of $10 \%$ nitrogen and $6 \%$ phosphoric acid and $F_2$ consists of $5 \%$ nitrogen and $10 \%$ phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast $14 \ kg$ of nitrogen and $14 \ kg$ of phosphoric acid for her crop. If $F _1$ costs Rs $6 / kg$ and $F _2$ costs $Rs 5 / kg$, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?

If $\text{x}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)\text{ and y}=\text{a}\Big(\text{t}-\frac{1}{\text{t}}\Big),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$

Show that $\text{A}=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ sastisfies the equation $A^2 + 4A - 42I = 0.$ Hence find $A^{-1}.$

Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$

Find the equation of the plane which contains two parallel lines $\frac{\text{x}-4}{1}=\frac{\text{y}-3}{-4}=\frac{\text{z}-2}{5}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-4}=\frac{\text{z}}{5}.$