Question
For each of the differential equations given in find the general solution: $\cos^2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}\Big(0\leq\text{x}<\frac{\pi}{2}\Big)$
✓
Answer
Given: Differential equation $\cos^2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\tan\text{x}$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos^2\text{x}}=\frac{\tan\text{x}}{\cos^2\text{x}}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}+(\sec^2\text{x})\text{y}=\sec^2\text{x}\tan\text{x}$
$\text{Comparing with}\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{we have P}=\sec^2\text{x and Q}=\sec^2\text{x}\tan\text{x}.$
$\therefore\ \ \int\text{P dx}=\int\sec^2\text{x dx}=\tan\text{x}\ \ \text{I.F}=\text{e}^{\int\text{P dx}}=\text{e}^{\tan\text{x}}$
$\text{Solution is y (I.F)}=\int\text{Q (I.F.) dx}+\text{c}$ $\Rightarrow\ \ \text{ye}^{\tan\text{x}}=\int\sec^2\text{x}\tan\text{xe}^{\tan\text{x}}\text{dx}+\text{c}\ \ ...\text{(i)}$
$\text{Putting}\tan\text{x}=\text{t and differentiating}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\ \ \int\sec^2\text{x}\ \tan\text{xe}^{\tan\text{x}}\text{ dx}=\int\text{te}^\text{t}\ \text{dt}$
Applying product rule,
$\Rightarrow\ \ \int\sec^2\text{x}\tan\text{x e}^{\tan\text{x}}$ $\text{dx}=\text{t.e}^\text{t}-\int1.\text{e}^\text{t}\ \text{dt}=\text{t.}\text{e}^\text{t}-\text{e}^\text{t}=(\text{t}-1)\text{e}^\text{t}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}$
Putting this value in eq. (i),
$\text{ye}^{\tan\text{x}}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{C}$ $\ \ \Rightarrow\ \ \text{y}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{ce}^{\tan\text{x}}$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos^2\text{x}}=\frac{\tan\text{x}}{\cos^2\text{x}}$ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}+(\sec^2\text{x})\text{y}=\sec^2\text{x}\tan\text{x}$
$\text{Comparing with}\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{we have P}=\sec^2\text{x and Q}=\sec^2\text{x}\tan\text{x}.$
$\therefore\ \ \int\text{P dx}=\int\sec^2\text{x dx}=\tan\text{x}\ \ \text{I.F}=\text{e}^{\int\text{P dx}}=\text{e}^{\tan\text{x}}$
$\text{Solution is y (I.F)}=\int\text{Q (I.F.) dx}+\text{c}$ $\Rightarrow\ \ \text{ye}^{\tan\text{x}}=\int\sec^2\text{x}\tan\text{xe}^{\tan\text{x}}\text{dx}+\text{c}\ \ ...\text{(i)}$
$\text{Putting}\tan\text{x}=\text{t and differentiating}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\ \ \int\sec^2\text{x}\ \tan\text{xe}^{\tan\text{x}}\text{ dx}=\int\text{te}^\text{t}\ \text{dt}$
Applying product rule,
$\Rightarrow\ \ \int\sec^2\text{x}\tan\text{x e}^{\tan\text{x}}$ $\text{dx}=\text{t.e}^\text{t}-\int1.\text{e}^\text{t}\ \text{dt}=\text{t.}\text{e}^\text{t}-\text{e}^\text{t}=(\text{t}-1)\text{e}^\text{t}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}$
Putting this value in eq. (i),
$\text{ye}^{\tan\text{x}}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{C}$ $\ \ \Rightarrow\ \ \text{y}=(\tan\text{x}-1)\text{e}^{\tan\text{x}}+\text{ce}^{\tan\text{x}}$
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