Solve the following initial value problems: dy dx +2y=e^ -2x , y(… - Vidyadip

Question

Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{e}^{-2\text{x}}\sin\text{x},\text{ y}(0)=0$

✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{e}^{-2\text{x}}\sin\text{x}\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=2$ and $\text{Q}=\text{e}^{-2\text{x}}\sin\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\text{e}^-{2\text{x}}\sin\text{x}$
$\Rightarrow\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\int\sin\text{x dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+\text{C}\ ....(\text{ii})$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^0=-\cos0+\text{C}$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}\text{e}^{2\text{x}}=-\cos\text{x}+1$
$\Rightarrow\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$
Hence, $\text{y}\text{e}^{2\text{x}}=1-\cos\text{x}$ 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 "5 Marks Questions" from Differential Equations→Differential Equations — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Integrate the function: $\frac{1}{1+\cot x}$

Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x},\text{ y}(0)=1$

Evaluate the following integrals: $\int\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}\ \text{dx}$

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}\ \text{dx}$

Obtain the Inverse of the following matrix using elementary operations:$A= \begin{vmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{vmatrix}.$

Evaluate :
$\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, -4) and is perpendicular to the line with direction ratios 2, 3, -1.

Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.

If $a, b, c$ are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either $a + b + c = 0$ or $a = b= c.$