Download App

Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0,\text{y}(0)=1,\text{y}(0)=3$
Function $\text{y}=\text{e}^\text{x}+\text{e}^{2\text{x}}$
✓

Answer

$\text{y}=\text{e}^{\text{x}}+\text{e}^{2\text{x}} ...(\text{i})$

Differentiating it with respect to x,

$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}+2\text{e}^{2\text{x}} ...\text{(ii)}$

Again, differentiating it with respect to x,

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{e}^{\text{x}}+4\text{e}^{2\text{x}}$

$=(3-2)\text{e}^{\text{x}}+(6-2)\text{e}^{2\text{x}}$

$=3\text{e}^{\text{x}}+6\text{e}^{2\text{x}}-2\text{e}^{\text{x}}-2\text{e}^{2\text{x}}$

$=3(\text{e}^\text{x}+2\text{e}^{2\text{x}})-2 (\text{e}^{\text{x}}+\text{e}^{2\text{x}})$

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=3\frac{\text{dy}}{\text{dx}}-2\text{y}$

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$

It is the given equation, so

 y - ex + 2e2x is the solution of the given equation.

put x = 0 in equation (i),

y = e0 + e0

y = 1 + 1

y = 2

so,

y(0) = 2

put x - 0 in equation (ii),

$\frac{\text{dy}}{\text{dx}}=\text{e}^{0}+2\text{e}^{0}$

y' = 1 + 2

y' = 3

so,

y'(0) = 3

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 "4 Marks" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}$
→
Evaluate the following integrals:
$\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}$
→
Evaluate the follwing intregals:
$\int\frac{1}{\text{x}(\text{x}^4-1)}\ \text{dx}$
→
Solve the following differential equations:
$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
→
If $\vec{\text{a}} \times \vec{\text{b}} = \vec{\text{c}} \times \vec{\text{d}}$ and $\vec{\text{a}} \times \vec{\text{c}} = \vec{\text{b}} \times \vec{\text{d}},$ show that $\vec{\text{a}} - \vec{\text{d}}$ is parallel to $\vec{\text{b}} - \vec{\text{c}},$ where $\vec{\text{a}} \neq \vec{\text{d}}$ and $\vec{\text{b}} \neq \vec{\text{c}}.$
→
If $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors inclined at an angle $\theta$, prove that

$\cos\frac{\theta}{2}=\frac{1}{2}\big|\hat{\text{a}}+\hat{\text{b}}\big|$

→
Evaluate: $\int\frac{\sin(\text{x} - \text{a})}{\sin(\text{x + a})}\text{ dx}.$
→
Solve the following initial value problems:
$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
→
One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
y = x4 - 6x3 + 13x2 - 10x + 5 at x = 1
→