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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
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Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
Integrating both sides, we get
$\int\frac{\text{dy}}{\text{dx}}=\int(\cos^3\text{x}\sin^2\text{x}+\text{x}\sqrt{2\text{x}+1})\text{dx}$
$\Rightarrow\text{y}=\int\cos^3\text{x}\sin^2\text{x dx}+\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
$\Rightarrow\text{y}=\text{I}_1+\text{I}_2\ ...(1)$
Where
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Now,
$\text{I}_1=\int\cos^3\text{x}\sin^2\text{x dx}$
$=\int\sin^2\text{x}(1-\sin^2\text{x})\cos\text{x dx}$
Putting $\text{t}=\sin\text{x},$ we get
$\text{dt}=\cos\text{x dx}$
$\Rightarrow\text{I}_1=\int\text{t}^2(1-\text{t}^2)\text{dt}$
$=\int(\text{t}^2-\text{t}^4)\text{dt}$
$=\frac{\text{t}^3}{3}-\frac{\text{t}^5}{5}+\text{C}_1$
$=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1$
$\text{I}_2=\int\text{x}\sqrt{2\text{x}+1}\text{dx}$
Putting $\text{t}^2=2\text{x}+1$ we get,
$2\text{t dt}=2\text{dx}$
$\Rightarrow\text{t dt}=\text{dx}$
Now,
$\text{I}_2=\int\Big(\frac{\text{t}^2-1}{2}\Big)\text{t}\times\text{t}\text{ dt}$
$=\frac{1}{2}\int(\text{t}^4-\text{t}^2)\text{dt}$
$=\frac{\text{t}^5}{10}-\frac{\text{t}^3}{6}+\text{C}_2$
$=\frac{(2\text{x}+1)\frac{5}{2}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}_2$
Putting the value of $I_1$ and $I_2$ in (1), we get
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\text{C}_1+\frac{(2\text{x}+1)^{\frac{5}{2}}}{6}+\text{C}_2$
$\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$
Hence, $\text{y}=\frac{\sin^3\text{x}}{3}-\frac{\sin^5\text{x}}{5}+\frac{(2\text{x}+1)^{\frac{5}{2}}}{10}-\frac{(2\text{x}+1)^{\frac{3}{2}}}{6}+\text{C}$ is the solution to the given differential equation.

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